B- lonic Bonds- Some atoms have such a strong attraction for electrons.docx

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Some atoms have a strong attraction for electrons, called electronegativity, that allows them to pull electrons away from other atoms to form ions. Atoms that gain electrons become negatively charged anions, while atoms that lose electrons become positively charged cations. Cations and anions are attracted to each other due to their opposite charges and form ionic compounds. Ionic bonds result from the electrostatic attraction between oppositely charged ions.

B. lonic Bonds: Some atoms have such a strong attraction for electrons (called electronegativity)
that they will steal electrons that are loosely held in place from other atoms. The specific
structure of an atom will determine whether it gains or loses electrons to form ions. Ion are
charged atoms, and atoms electrons have a negative charge (one negative for each extra
electron), and atoms which have a positive charge are atoms which have lost electrons Positively
charged ions are called cations. Negatively charged ions are anions. Like charges repel, while
unlike charges attract (Opposites attract). Cations are then attracted to anions, and together, they
will from ionic compounds. 1. Using the periodic table of elements, analyze the following
element, ions and compounds. You will be asked to determine the total number of electrons ro
protons in the atom or molecule as written. a. Protons in Na b Protons in Na* c. Protons in Cl d.
Protons in Cr e. Protons in Ca f. Protons in Ca2 g. Protons in O h. Protons in O2 i. Protons in
CaO k. Electrons in NacI m. Electrons in Na* j. Electrons in Cao I. Electrons in Na n. Electrons
in Cl o. Electrons in C p. Electrons in Ca q. Electrons in Ca r. Electrons in O . Electrons in O2
Solution
a. protons in Na = 11
b. protons in Na +
= 11
c. protons in Cl = 17
d. protons in Cl -
= 17
e. protons in Ca = 20
f. protons in Ca +2
= 20
g. protons in O = 8
h. protons in O -2
= 8
i. protons in CaO = 28
j. electrons in CaO = 28

K. protons in NaCl = 28
L. electrons in Na = 11
m. electrons in Na +
= 10
n. electrons in Cl = 17
o. electrons in Cl -
= 16
p. electrons in Ca = 20
q. electrons in Ca +2
= 18
r. electrons in O = 8
s. electrons in O -2
= 10

Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2? Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3800 Character(s) remaining Solution The atom with less electronegative is central atom and the atoms with more electronegtive are terminal atoms. So, Sulphur with less electronegative is central atom and Oxygens are terminal atoms. S ( Z = 16 ) 1st excited state electronic configuration is, 1s2 s2 2p6 3s2 3p3 3d1 S in its first exicted state undergoes sp2 hybirdisation due ot which it has 1 paired sp2 hybrid orbital and 2 unpaired sp2 hybrid orbitals. And also S has one p and one d orbital with unpaired electrons. O (Z = 8) E.C is 1s2 2s2 2p4 Each O has two unpaired electrons in its valence shell. The each sp2 unpaired hybrid orbital of S overlaps with one of the unpaired p - orbital ofeach oxygen along the internuclear axis to form single S - O sigma bond with each Oxygen. The reamining unpaired pure p-orbitals of S overlaps with unpaired p - orbital of each oxygen perpendicular to the internuclear axis to form one pi bonds with each oxygen. Still S ia remained with one lone pair of electons. Due to this SO 2 is in bent shape and has the bond angle < 120 0 .

Balance the redox reaction by inserting the appropriate coefficients-.docx

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Balance the redox reaction by inserting the appropriate coefficients. redox reaction: | Fe3+ + NO + H2O- Fe2+ + H+ + No; Solution Fe in Fe+3 has oxidation state of +3 Fe in Fe+2 has oxidation state of +2 So, Fe in Fe+3 is reduced to Fe+2 N in NO2- has oxidation state of +3 N in NO3- has oxidation state of +5 So, N in NO2- is oxidised to NO3- Reduction half cell: Fe+3 + 1e- --> Fe+2 Oxidation half cell: NO2- --> NO3- + 2e- Balance number of electrons to be same in both half reactions Reduction half cell: 2 Fe+3 + 2e- --> 2 Fe+2 Oxidation half cell: NO2- --> NO3- + 2e- Lets combine both the reactions. 2 Fe+3 + NO2- --> 2 Fe+2 + NO3- Balance Oxygen by adding water 2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- Balance Hydrogen by adding H+ 2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- + 2 H+ This is balanced chemical equation in acidic medium Answer: 2 Fe 3+ + NO 2 - + H 2 O â€”> 2 Fe 2+ + 2 H + + NO 3 - .

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