Important Questions for CSIR NET Chemical Science 2022 - Inorganic Chemistry-II

Most Important Questions On Inorganic Chemistry- II 

1. For boranes, the correct order of acidity is given by:

1. Closo > Nido > Arachno
2. Arachno > Nido > Closo
3. Closo > Arachno > Nido
4. Arachno > Closo > Nido

2. For Cubane-ferredoxin,

The value of n, m and p respectively are?

1. 4,8,4
2. 4,4,4
3. 4,8,8
4. 2,4,4

3. What will be the symmetry of the anti-bonding molecular which is formed by a linear combination of the px or py atomic orbitals in a homonuclear diatomic molecule? 

4. Give the order for bond energy for the following-

1. NH-NH > F-F > O-O
2. NH-NH > O-O > F-F
3. O-O > F-F > NH-NH
4. F-F < O-O < NH-NH

5. The lowest energy term for low spin d5 is:

1. 6S
2. 2I
3. 2H
4. 6F

6. Which one of the following is coloured in an aqueous medium?

1. La3+
2. Lu3+
3. Gd3+
4. Eu3+

7. Which of the following statements is incorrect about Mossbauer spectroscopy?

1. It is also known as NGR.
2. It is used to estimate the oxidation state of metal.
3. This does not give any information about the symmetry of molecules.
4. This spectroscopy is applicable to Au, Sn, Fe.

8. The correct sequence for reactivity of interhalogens is:

1. ClF < BrF
2. BrF > IF
3. BrF5 > ClF3
4. . IF5 > IF7

9. Calculate the Bonding molecular orbital of the following cage structure [Fe4(CO)12C]2-.

1. 35
2. 32
3. 41
4. 31

10. Arrange the following metal with respect to the increase in melting point Cr, Mo, and W.

1. W< Mo > Cr
2. W< Mo <Cr
3. W> Mo > Cr
4. W> Cr > Mo

Answer Key:

1. B
2. C
3. C
4. D
5. B
6. D
7. C
8. B
9. D
10. C

Solutions:

Solution 1. 

In the case of the arachnoid, two apexes are missing, whereas, in case of Nido, one apex is missing. And Closo is a polyhedron. Thus, arachnoid can easily accept electrons as compared to both Nido and closo. Hence, the correct answer is (B).

Solution 4. 

More the number of lone pairs more will be the interelectronic repulsion and less will be the bond energy. Hence, the order becomes:

F-F < O-O < NH-NH

Solution 5.

Therefore, the lowest energy term of low spin d5 is 2I.

Solution 6. f-block elements having half-filled, zero or fully filled orbitals are not coloured in an aqueous medium whereas those which are partially filled are coloured in an aqueous medium.

La3+:  4f0

Lu3+:  4f14

Gd3+: 4f7

Eu3+:  4f6

Solution 7. Mossbauer spectroscopy is also known as NGR (nuclear gamma resonance) since this spectroscopy uses gamma radiations. It also helps to determine the oxidation state, as more the oxidation state, less will be electron density then this will affect the isomer shift (can have an inverse or direct relation, depending on the molecule). This gives information about the symmetry of molecules. If the symmetry (electronic and ligand both) of the molecule is spherical, then electric field gradient is zero and quadrupole splitting will be absent, whereas if symmetry is non-spherical, then electric field gradient is nonzero and quadrupole splitting will be present. This spectroscopy is applicable to Au, Sn, Fe.

Solution 8. For interhalogen, the reactivity depends on the electronegativity of the halogen bond.  More the electronegativity difference less will be the reactivity. If an interhalogen contains the same halogen atom, out of which one is F, then it also depends on the number of F atoms attached. The more the number of F atoms more will be the reactivity. Thus, the correct option is (B) as the electronegativity difference between Br and F is smaller as compared to I and F.

Solution 9. Determination of Bonding molecular orbital:-

Step I → Calculate TVE (Total Valence e- count.)

Step II → Add the value of the valence electron of interstitial atoms.

Eg. H → 1 e-

H2 → 2 e-

Carbon family → 4 e-

Nitrogen family → 5 e-

Step III →Calculate Bonding molecular orbital

BMO = (TVE )/2

Calculate T.V.E as:

C = 4 e- Contribution because it is encapsulated atom

Os = 3d64s2 = 8 e- contribution

= 4+(8 x 4) + (12 x 2) +2

= 62 e-

BMO = Total valence electron /2

=62/2

= 31

Solution 10. This is because the d- block elements have a high number of unpaired electrons which contributes to a greater number of electrons in the crystal lattice thereby increasing the strength of the metallic bond between the atoms. Therefore, the Melting and Boiling points are high. And on moving down the period, the density of metal increases due to lanthanide contraction. Hence, the order of melting point of given metals are:

 W> Mo > Cr

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