Inorganic chemistry

1. Answer Key
F.Y.B.Sc. Ist SEMESTER END EXAMINATION, NOVEMBER 2022
Subject: Chemistry
Paper code and Name: CHC101 Inorganic Chemistry and Organic Chemistry
Time Duration: 2 Hours Total Marks: 80
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Section A: Inorganic Chemistry-1 Marks: 40
Q.1. Answer any five from the following (2 x 5 = 10 Marks)
i. Draw the pictures of 1s and 2s orbitals in the XY-plane as flat projections in the plane of the
paper.
ii. Write the ground state electronic configuration of a) Be+
and b) F.
a) Be (Z=4) =1s2
,2s2
Be+
in ground state 1s2
, 2s1
b) F (Z=9) =1s2
, 2s1
, 2p5
,
F in ground state 1s2
, 2s1
, 2p5
,
iii. Describe the general form of the time-independent Schrodinger wave equation?

2. iv. A covalent molecule have a definite molecular shape, while an ionic molecule does not.
Why?
A covalent bond is formed by sharing of electrons thus, covalent bonds are directional. Atoms are
bonded in specific orientations relative to one another. This gives molecules definite shapes.
However ionic bond is nondirectional as the association between the pair of ions (cation and anion)
is considered purely electrostatic.
v. Give any four general characteristics of ionic bonding.
The compounds that exhibit ionic bonds show the following characteristics
• high melting points.
• hard and brittle.
• dissociate into ions when dissolved in water.
• Pure solid materials do not conduct electricity.
• Molten and in-solutions ionic compounds conduct electricity.
vi. Why pi bond is considered weaker than the sigma bond?
The Sigma bond is formed by axial overlap while the Pi bond is formed by lateral (sideways)
overlap. Thus, in the sigma bond, the overlapping is maximum whereas in the pi bond the
overlapping is lesser comparatively. As a result, the pi bond is weaker than the sigma bond.
vii. Predict the structures of CO2 and H2O molecules.
Q.2. A. Answer the following
i) Calculate the bond order of peroxide molecular ion. 4 Marks

3. ii) Plot the radial wave function for 3s, 3p, and 3d orbitals. 3 Marks
OR
Q.2. A. iii) Draw the Born-Haber cycle for the KCl molecule. 4 Marks

4. iv) Write the Lewis structure of SiCl4, NaCl and nitrogen molecule 3 Marks
Q.2.B.i) Explain the anomalous electronic configuration of the Cr atom in the ground state.
4 Marks
Chromium (Cr) has an atomic number of 24. ·
Its expected electronic configuration is 1s2
,2s2
, 2p6
,3s2
, 3p6
, 4s2
, 3d4
Or [Ar] 3d4
, 4s2
However, chromium displays an anomalous configuration – [Ar] 3d5
, 4s1
.
This is because extra stability of half filled electronic configuration.

5. Hence chromium chooses to adopt a configuration in which exchange energy is maximized.
ii) Compare the polarization and polarizability of I and F atoms in HI and HF molecules
respectively? 4 Marks
Polarisation- The negative cloud of electrons is distorted due to the influence of positive atomic
nuclei.
Polarisability- the ability of an atom/ion to be distorted due to the influence of an electric field
such as that of an atom.
Polarisability is directly proportional to the size of the anion. Fluorine is very small compared to
iodine atoms. As hydrogen is small and electropositive compared to both halogen atoms the
positive charge on it will polarize the iodine atom to a greater extent than the fluorine atom.
Q.3. A. Answer the following.
i) Account for the change in the bond angle of NH3 and NH4
+
molecules. 4 Marks
In both the molecules the N is sp3
hybridized. In the NH4
+
ion, all four sp3
hybridized orbitals are
bonded whereas in NH3 there is a lone pair on N, which is responsible for lone pair-bond pair
repulsion in NH3 reducing the bond angle from 109.5∘ to 107∘
ii) How many radial and angular nodes do 1s, 2s, and 3s orbitals exhibit? 3 Marks

6. 1s-orbital has no (zero) radial node and angular node (n=1, l=0)
2s-orbital has one radial node but no angular node (n=2, l =0)
3s-orbital has two radial nodes but no angular node (n=3, l =0)
Number of radial nodes = (n - l) - 1
OR
Q.3. A. iii) Comment on the trend observed in the stability of alkaline earth metal carbonates.
4 Marks
The alkaline earth metal carbonates are BeCO3, MgCO3, CaCO3, SrCO3, and BaCO3.
The (CO3)2-
is a bigger anion that stabilizes bigger cations. The thermal decomposition for the
metal carbonates is favored for lower size cations as smaller oxide ions stabilize the by higher
lattice energy compensation.
The above figure represents a thermodynamic cycle showing the enthalpy changes involved in the
decomposition of a solid carbonate.
MCO3 (s)
→ MO(s)
+ CO2 (g)
ΔHdecom MCO3(s)
The higher the lattice energy for metal oxide (ΔHL MO(s)) lower will be the decomposition enthalpy
(ΔHdec MCO3(s)) for the metal carbonate.
Lattice energy is inversely proportional to internuclear distance. As the size of the cation down the
group increases the lattice enthalpy for the formation of metal oxide decreases.
Down the group as ionic size increases, the carbonates require more heating to decompose. This
means that alkaline earth carbonates become more thermally stable down the group.
iv) Describe the spin quantum number and magnetic spin quantum number. 3 Marks
The Spin Quantum Number (s) describes the angular momentum of an electron. (s=1/2)

7. The Magnetic Spin Quantum number (ms) gives information about the direction of spinning of
the electron present in any orbital. (ms = + ½ or - ½ )
Q.3.B i) Explain the dual behavior of matter. 4 Marks
Rutherford and Bohr describe the atom as a central nucleus surrounded by electrons in certain
orbits. However, according to the Heisenberg uncertainty principle, it is difficult to find the exact
location and momentum associated with such small subatomic particles. de Broglie proposed that
matter should also exhibit dual behavior i.e. both particle and wave-like properties. He showed
that the wavelength associated with large bodies is negligible and hence not significant.
𝜆=ℎ/𝑚𝑣
𝜆= wavelength, h= planks constant, m=mass, v =velocity
Different theories support the dual nature of the electron.
Particle- Einstein Photoelectric effect.
Wave- interference, diffraction,
ii) Calculate the percentage of ionic character for CO molecule. 4 Marks
(Given: Dipole moment is 0.37 X 10-30
C m; charge on electron e = 1.60210 X 10-19
C; bond
length = 112.8 pm)
bond length = 112.8 pm= 112.8 X10-12
m
Charge (Q) = =
𝐷𝑖𝑝𝑜𝑙𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 (𝜇)
𝑏𝑜𝑛𝑑 𝑙𝑒𝑛𝑔𝑡ℎ(𝑟)
=
0.37 × 10−30 C m
112.8 × 10−12 m
= 3.2801 × 10−21
C
% ionic character = =
𝐶ℎ𝑎𝑟𝑔𝑒 (𝑄)
charge on electron (𝑒)
× 100
=
3.2801 × 10−21
C
1.60210 × 10−19 C
× 100
= 2.05%

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