PRACTICE WORKSHEET VBT

 PRACTICE WORKSHEET

Class 12 Chemistry – Coordination Compounds

VBT – 10 Questions

1. Predict the hybridisation, geometry and magnetic nature of $[Co(NH_3)_6]^{3+}$.

2. Predict the hybridisation, geometry and magnetic nature of $[CoF_6]^{3-}$.

3. Determine the hybridisation and magnetic nature of $[Fe(CN)_6]^{4-}$.

4. Determine the hybridisation and magnetic nature of $[FeF_6]^{3-}$.

5. Predict the hybridisation, geometry and magnetic nature of $[Ni(CN)_4]^{2-}$.

6. Predict the hybridisation, geometry and magnetic nature of $[NiCl_4]^{2-}$.

7. Explain, using VBT, why $[Ni(CN)_4]^{2-}$ is diamagnetic while $[NiCl_4]^{2-}$ is paramagnetic.

8. Determine the hybridisation, geometry and number of unpaired electrons in $[CoF_6]^{3-}$.

9. Determine the hybridisation, geometry and magnetic nature of $[Fe(CN)_6]^{3-}$.

10. Compare $[Co(NH_3)_6]^{3+}$ and $[CoF_6]^{3-}$ on the basis of hybridisation, geometry and magnetic behaviour.

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ANSWER

Class 12 Chemistry – Coordination Compounds

VBT (Valence Bond Theory) – 10 Questions with Solutions

1. Predict the hybridisation and geometry of $[Co(NH_3)_6]^{3+}$.

Solution:
Co = 27
Co³⁺ = $3d^6$

$NH_3$ acts as a strong enough ligand here to cause pairing of 3d electrons.

Hybridisation = $d^2sp^3$

Geometry = Octahedral

Answer: $d^2sp^3$, octahedral, inner-orbital complex.

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2. Predict the hybridisation and geometry of $[CoF_6]^{3-}$.

Solution:
Co³⁺ = $3d^6$

$F^-$ is a weak-field ligand, so pairing does not occur.

Hybridisation = $sp^3d^2$

Geometry = Octahedral

Answer: $sp^3d^2$, octahedral, outer-orbital complex.

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3. Determine the hybridisation and magnetic nature of $[Fe(CN)_6]^{4-}$.

Solution:
Fe²⁺ = $3d^6$

$CN^-$ is a strong-field ligand and causes pairing.

Hybridisation = $d^2sp^3$

All electrons are paired.

Answer: $d^2sp^3$, octahedral, diamagnetic.

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4. Determine the hybridisation and magnetic nature of $[FeF_6]^{3-}$.

Solution:
Fe³⁺ = $3d^5$

$F^-$ is a weak-field ligand, so no pairing occurs.

Hybridisation = $sp^3d^2$

There are 5 unpaired electrons.

Answer: $sp^3d^2$, octahedral, paramagnetic (5 unpaired electrons).

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5. Determine the hybridisation and geometry of $[Ni(CN)_4]^{2-}$.

Solution:
Ni²⁺ = $3d^8$

$CN^-$ is a strong-field ligand and causes pairing of 3d electrons.

Hybridisation = $dsp^2$

Geometry = Square planar

All electrons are paired.

Answer: $dsp^2$, square planar, diamagnetic.

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6. Determine the hybridisation and magnetic nature of $[NiCl_4]^{2-}$.

Solution:
Ni²⁺ = $3d^8$

$Cl^-$ is a weak-field ligand, so pairing does not occur.

Hybridisation = $sp^3$

Geometry = Tetrahedral

There are 2 unpaired electrons.

Answer: $sp^3$, tetrahedral, paramagnetic (2 unpaired electrons).

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7. Why is $[Ni(CN)_4]^{2-}$ diamagnetic whereas $[NiCl_4]^{2-}$ is paramagnetic?

Solution:
Both contain Ni²⁺ ($3d^8$).

• $CN^-$ is a strong-field ligand → pairing occurs → $dsp^2$ hybridisation → no unpaired electrons.

• $Cl^-$ is a weak-field ligand → no pairing → $sp^3$ hybridisation → 2 unpaired electrons.

Answer: Difference in ligand strength causes different hybridisation and magnetic behaviour.

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8. Predict the hybridisation, geometry and magnetic nature of $[CoF_6]^{3-}$.

Solution:
Co³⁺ = $3d^6$

$F^-$ is weak-field → no pairing.

Hybridisation = $sp^3d^2$
Geometry = Octahedral
Unpaired electrons = 4

Answer: $sp^3d^2$, octahedral, paramagnetic (4 unpaired electrons).

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9. Determine the hybridisation and magnetic nature of $[Fe(CN)_6]^{3-}$.

Solution:
Fe³⁺ = $3d^5$

$CN^-$ is strong-field → pairing occurs.

Configuration after pairing:

$$(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow)(\uparrow)(\uparrow)$$

Two 3d orbitals become vacant.

Hybridisation = $d^2sp^3$

Unpaired electrons = 1

Answer: $d^2sp^3$, octahedral, paramagnetic (1 unpaired electron).

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10. Compare $[Co(NH_3)_6]^{3+}$ and $[CoF_6]^{3-}$ on the basis of VBT.

Solution:

Property	$[Co(NH_3)_6]^{3+}$	$[CoF_6]^{3-}$
Metal ion	Co³⁺	Co³⁺
$d$-configuration	$d^6$	$d^6$
Ligand	$NH_3$	$F^-$
Ligand strength	Stronger	Weak
Pairing	Occurs	Does not occur
Hybridisation	$d^2sp^3$	$sp^3d^2$
Geometry	Octahedral	Octahedral
Unpaired electrons	0	4
Magnetic nature	Diamagnetic	Paramagnetic

Key VBT rule:
Strong-field ligand → pairing → inner orbital $d^2sp^3/dsp^2$
Weak-field ligand → no pairing → outer orbital $s{p}^3{d}^2\mathrm{/}s{p}^3$