This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
\"ligands\" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See \"spectrochemical series\" for more detail about this. 6. To find
\"exact\" color of metal ions, you need a supercomputer to do so. (I\'m not sure whether today
computer can calculate it). So, it\'s better to memorize the colors. Personally, I think you\'ll
understand about metal\'s color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends .
In this presentation I cover all of the HL content for T13- Periodic trends. As usual I would like to acknowledge the Pearson textbook as the main source of my inspiration.
In this presentation I cover all of the HL content for T13- Periodic trends. As usual I would like to acknowledge the Pearson textbook as the main source of my inspiration.
How do we describe the bonding between transition metal (ions) and their ligands (like water, ammonia, CO etc) ?
The Crystal Field Model gives a simple theory to explain electronic spectra.
The bonding in lanthanide complexes differs from that seen in transiti.docxchrisflorence13710
The bonding in lanthanide complexes differs from that seen in transition metal complexes. Discuss how this difference in bonding explains why for a transition metal ion, such as Co3+, the colour of its complexes varies depending on the co-ordinated ligands, whilst all complexes of a particular lanthanide ion, such as Eu3+, exhibit a similar colour (b)
Solution
Lanthanides (inner transition metals) resemble 3d transition metals mainly in forming coloured complexes.
But they differ in the oxidation state which is due to large reduction potential for oxidation state when related to 1 electron reduction in case transition compounds.
There is a lack in 4f chemistry of very high oxidation state is familiar to 3d transition complexes. The 4f metals are not willing to form strong Ln = O double bond complexes and lack of significant third ionization energy IE 3 is important in maintaining +3 oxidation state at the end of the lanthanide series.
The increase in atomic radius is greater than between 3d and 4d metals than between the 4d and 5d metals because of lanthanide contraction.
Sigma bonding interaction in coordination compound is due to attraction of electrons on the ligand for the charge of metal ion.
Electrons in ligands repel electrons in unhybridised d orbital of metal ion. The transition metal ion like Co +3 complexes exhibit many intense colours in host crystals or solutions.
The colour of light absorbed by complexed ion is related to electronic energy changes in structure of complex.
The electron leaping from a lower energy state to a higer energy state. If the crystal field strength is weak and the energy gap is small leading to unpaired electrons and form a paramagnetic complex in case of 4th and 5th electron going to higher energy d z 2 and d x 2 - d y 2 .
The fourth through 6 electrons will pair with d xy , d yz , and d zx . if the field is strong and the energy gap is large, leading to paired electrons and a diamagnetic complex
The colour of transition metal ion, Co +3 complexes differs depends on the coordinated ligands, which is due to change in electronic energy in the structure of complex while colour of particular lanthainde Eu 3+ (europium) exhibit similar pale pink clour this is due to f-->f transition rather than d--->d transition.
.
Class XII d and f- block elements (part 2)Arunesh Gupta
This part contains ionisation enthalpies, oxidation states, metal oxides & oxocations, magnetic properties, coloured ions of d block elements, catalysts, interstitial compounds, alloy formation & some important conceptual questions with answers as hints. Also some reasoning questions are given to test the understanding of properties of d block elements.
How do we describe the bonding between transition metal (ions) and their ligands (like water, ammonia, CO etc) ?
The Crystal Field Model gives a simple theory to explain electronic spectra.
The bonding in lanthanide complexes differs from that seen in transiti.docxchrisflorence13710
The bonding in lanthanide complexes differs from that seen in transition metal complexes. Discuss how this difference in bonding explains why for a transition metal ion, such as Co3+, the colour of its complexes varies depending on the co-ordinated ligands, whilst all complexes of a particular lanthanide ion, such as Eu3+, exhibit a similar colour (b)
Solution
Lanthanides (inner transition metals) resemble 3d transition metals mainly in forming coloured complexes.
But they differ in the oxidation state which is due to large reduction potential for oxidation state when related to 1 electron reduction in case transition compounds.
There is a lack in 4f chemistry of very high oxidation state is familiar to 3d transition complexes. The 4f metals are not willing to form strong Ln = O double bond complexes and lack of significant third ionization energy IE 3 is important in maintaining +3 oxidation state at the end of the lanthanide series.
The increase in atomic radius is greater than between 3d and 4d metals than between the 4d and 5d metals because of lanthanide contraction.
Sigma bonding interaction in coordination compound is due to attraction of electrons on the ligand for the charge of metal ion.
Electrons in ligands repel electrons in unhybridised d orbital of metal ion. The transition metal ion like Co +3 complexes exhibit many intense colours in host crystals or solutions.
The colour of light absorbed by complexed ion is related to electronic energy changes in structure of complex.
The electron leaping from a lower energy state to a higer energy state. If the crystal field strength is weak and the energy gap is small leading to unpaired electrons and form a paramagnetic complex in case of 4th and 5th electron going to higher energy d z 2 and d x 2 - d y 2 .
The fourth through 6 electrons will pair with d xy , d yz , and d zx . if the field is strong and the energy gap is large, leading to paired electrons and a diamagnetic complex
The colour of transition metal ion, Co +3 complexes differs depends on the coordinated ligands, which is due to change in electronic energy in the structure of complex while colour of particular lanthainde Eu 3+ (europium) exhibit similar pale pink clour this is due to f-->f transition rather than d--->d transition.
.
Class XII d and f- block elements (part 2)Arunesh Gupta
This part contains ionisation enthalpies, oxidation states, metal oxides & oxocations, magnetic properties, coloured ions of d block elements, catalysts, interstitial compounds, alloy formation & some important conceptual questions with answers as hints. Also some reasoning questions are given to test the understanding of properties of d block elements.
Similar to This is quite a broad question. I will try to acc.pdf (20)
VSEPR is easy if you follow the right steps. 1. .pdfrajat630669
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electron groups around the central
ion, so the shape is trigonal bipyramid if you draw in the lone pairs (trigonal plane with 2 groups
above and below the plane. You can determine that these two groups must be the lone pairs since
they repel more than the I-Cl bonds). Removing the lone pairs gives you a trigonal planar
molecule. The number of groups around the central ion determines the shape. 2 is linear 3 is
trigonal planar
Solution
VSEPR is easy if you follow the right steps. 1. Find the number of valence (outer)
electrons around the central atom (remember the charge) 2. Work out how many of these are
involved in bonding (remember you may have single or double bonds) 3. Remember the lone
pairs. 4. Work out the shape from the number of atoms bonded to the central atom and the
number of lone pairs. So for SeO2, Se is in group 6, and has 6 valence electrons. Se=O is a
double bond, so there are 2 of Se\'s electrons in a double bond There are two Se=O bonds, so 4
electrons are involved in Se=O bonds. This leaves another 2 electrons which are a lone pair. So
overall you have three groups around the central atom (2 Se=O bonds and a lone pair) so it is
trigonal planar if you drew in the lone pair. Removing the lone pair gives you a bent molecule.
For PF4+: P has 5 valence electrons, but it is a positive ion, so you knock off an electron to get 4
P-F bonds are single bonds, and there are 4 of them, which accounts for all the electrons.
Therefore PF4+ is a tetrahedral shape. Finally, ICl3: I has 7 valence electrons. I-Cl bonds are
single bonds, so the I uses one electron in each bond. Therefore 3 of the 7 electrons are involved
in I-Cl bonds, the other 4 must be in 2 lone pairs. This gives 5 electro.
What is an example of big data either from your personal experience .pdfrajat630669
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance information of the
aircraft.
Social Media Data : Social media such as Facebook and Twitter hold information and the views
posted by millions of people across the globe.
Stock Exchange Data : The stock exchange data holds information about the ‘buy’ and ‘sell’
decisions made on a share of different companies made by the customers.
Power Grid Data : The power grid data holds information consumed by a particular node with
respect to a base station.
Transport Data : Transport data includes model, capacity, distance and availability of a vehicle.
Search Engine Data : Search engines retrieve lots of data from different databases.
How is it similar or different to the example in the case? In response to your peers’ posts, discuss
how big data in the examples provided impact an organization in terms of the advantages and
disadvantages?
Big data is really critical to our life and its emerging as one of the most important technologies in
modern world. Follow are just few benefits which are very much known to all of us:
Using the information kept in the social network like Facebook, the marketing agencies are
learning about the response for their campaigns, promotions, and other advertising mediums.
Using the information in the social media like preferences and product perception of their
consumers, product companies and retail organizations are planning their production.
Disadvantages of Big Data:
- Unknown population representation
- Issues of data quality
- Typically not very multivariate (at the person level)
- Privacy and confidentiality issues
- Difficult to assess accuracy and uncertainty
Solution
What is an example of big data either from your personal experience or outside research?
Answer:
Big Data: Big data means really a big data, it is a collection of large datasets that cannot be
processed using traditional computing techniques. Big data is not merely a data, rather it has
become a complete subject, which involves various tools, technqiues and frameworks.
Big data involves the data produced by different devices and applications. Given below are some
of the fields that come under the Big Data.
Black Box Data : It is a component of helicopter, airplanes, and jets, etc. It captures voices of the
flight crew, recordings of microphones and earphones, and the performance.
S-Se=S with 3 lone pairs on the first S, 1 lone.pdfrajat630669
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S
Solution
S-Se=S with 3 lone pairs on the first S, 1 lone pair on the Se, and 2 lone pairs on
the second S.
The sample program for above series in JAVA will be like belowimpo.pdfrajat630669
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
D. This element can only have a -2 oxidation stat.pdfrajat630669
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2
Solution
D. This element can only have a -2 oxidation state in covalent compounds. is false.
note: such as S can be +4 in SO2.
solAn object will allocated statically when that object is needed.pdfrajat630669
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap
Solution
sol:
An object will allocated statically when that object is needed for entire life time of that program
1)if object needs recusrion and storage of object conservly then it is stored in stack
2)f object needs dynamic storage then it should be allocated in heap.
CO2 is non polar because C has no lone pairs, so .pdfrajat630669
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar.
Solution
CO2 is non polar because C has no lone pairs, so the structure is linear and hence
dipole cancell each other SO2 has lone pair, so the structure is bent due to the repulsion and
hence the structure has a net dipole moment, so polar..
As more halide ions are added, it loses its color.pdfrajat630669
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration.
Solution
As more halide ions are added, it loses its color and becomes more white due to the
ion concentration..
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public abstract class Appointment
{
private String description;
/**
Constructs an appointment without a description.
*/
public Appointment()
{
description = \"\";
}
/**
Sets the description of this appointment.
*/
public void setDescription(String description)
{
this.description = description;
}
/**
Determines if this appointment occurs on the given date.
*/
public abstract boolean occursOn(int year, int month, int day);
/**
Converts appointment to string description.
*/
public String toString()
{
return description;
}
}
Daily.java
public class Daily extends Appointment
{
public Daily (String description)
{
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
return true;
}
}
Onetime.java
public class Onetime extends Appointment
{
private int monthApp;
private int yearApp;
private int dayApp;
public Onetime(int yearAppInput, int monthAppInput, int dayAppInput, String description)
{
yearApp = yearAppInput;
monthApp = monthAppInput;
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if ((year == yearApp) && (month == monthApp) && (day == dayApp))
{
return true;
}
else
{
return false;
}
}
}
Monthly.java
public class Monthly extends Appointment
{
private int dayApp;
public Monthly (int dayAppInput, String description)
{
dayApp = dayAppInput;
setDescription(description);
}
public boolean occursOn(int year, int month, int day)
{
if (day == dayApp)
{
return true;
}
else
{
return false;
}
}
}
Solution
AppointmentDemo.java
import java.util.Scanner;
/**
Demonstration of the appointment classes
*/
public class AppointmentDemo
{
public static void main(String[] args)
{
Appointment[] appointments = new Appointment[4];
appointments[0] = new Daily(\"Brush your teeth.\");
appointments[1] = new Monthly(1, \"Visit grandma.\");
appointments[2] = new Onetime(2015, 11, 1, \"Dentist appointment.\");
appointments[3] = new Onetime(2015, 10, 31, \"Trick or Treat.\");
Scanner in = new Scanner(System.in);
int year = in.nextInt();
int month = in.nextInt();
int day = in.nextInt();
for (Appointment a : appointments)
{
if (a.occursOn(year, month, day))
{
System.out.println(a);
}
}
}
}
Appointment.java
/**
A class to keep track of an appointment.
*/
public ab.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
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This is quite a broad question. I will try to acc.pdf
1. This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
"ligands" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See "spectrochemical series" for more detail about this. 6. To find
"exact" color of metal ions, you need a supercomputer to do so. (I'm not sure whether today
computer can calculate it). So, it's better to memorize the colors. Personally, I think you'll
understand about metal's color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
"ligands" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See "spectrochemical series" for more detail about this. 6. To find
2. "exact" color of metal ions, you need a supercomputer to do so. (I'm not sure whether today
computer can calculate it). So, it's better to memorize the colors. Personally, I think you'll
understand about metal's color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
![This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
"ligands" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See "spectrochemical series" for more detail about this. 6. To find
"exact" color of metal ions, you need a supercomputer to do so. (I'm not sure whether today
computer can calculate it). So, it's better to memorize the colors. Personally, I think you'll
understand about metal's color when you study transition metals, like Crystal Field Theory, and
Ligand Field Theory.
Solution
This is quite a broad question. I will try to account all the important observations. 1.
Metal ions which do not have d-orbital electrons do not have color. (alkali, alkali-earth, Sc3+)
This is because of energy separation of s and p orbitals are not in the range of visible colors. 2.
Metal with fully-filled d orbitals do not have color. (Zn2+) This is because electrons in d-orbitals
cannot be excited to higher energy d-orbitals (all d-orbitals have the same energies, but when
influenced by other atom in complex, d-orbitals can split into different energy levels) 3. For ions
of the same metal, the ion with larger charge has a color of lower energy than the ion with
smaller charge. Example: Cr2+ (blue) and Cr3+ (green). This is because larger charge
contributes to larger split of the d-orbital. The more the orbitals split, the higher energy of light it
traps, allowing low-energy photon to enter our eyes. Note that you cannot be sure that the color
you see is a single color, not a combination of many colors. 4. For metals in the same column, a
metal ion which is below has a color of lower energy than an ion above. This can be explained
like no.3, larger ion causes more orbitals splitting. 5. Color of metal ions also depends on
"ligands" (complexing molecule/ion). e.g. Co2+ in water is pink [Co(H2O)6]2+, while Co2+ in
NH3 is blue [Co(NH3)4]2+. See "spectrochemical series" for more detail about this. 6. To find](https://image.slidesharecdn.com/thisisquiteabroadquestion-230412020910-4015e1d7/85/This-is-quite-a-broad-question-I-will-try-to-acc-pdf-1-320.jpg)
