In which one of the following pairs the central atoms exhibit sp^2 hybridization ?

Chemical Bonding and Molecular Structure PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q10 with Solution

View written solutionFree

Correct answer: A

Determine the hybridization of the central atom in each species

Hybridization depends on the steric number:

\text{Steric number} = \text{number of } \sigma\text{-bonds} + \text{number of lone pairs on central atom}

Steric number $= 3 \Rightarrow sp^2$
Steric number $= 4 \Rightarrow sp^3$

Check each molecule/ion

(i) $\mathrm{BF_3}$

Central atom: $\mathrm{B}$
Boron forms 3 $\sigma$ -bonds with 3 fluorine atoms
Lone pairs on B = 0

So,

\text{Steric number} = 3 + 0 = 3

Hence, central atom is $sp^2$ hybridized.

(ii) $\mathrm{NO_2^-}$

Central atom: $\mathrm{N}$
Nitrogen is bonded to two oxygens
Due to resonance, there are 2 $\sigma$ -bonds and 1 lone pair on N

So,

\text{Steric number} = 2 + 1 = 3

Hence, central atom is $sp^2$ hybridized.

(iii) $\mathrm{NH_2^-}$

Central atom: $\mathrm{N}$
Nitrogen forms 2 $\sigma$ -bonds with H atoms
Total valence electrons on N including negative charge give 2 lone pairs

So,

\text{Steric number} = 2 + 2 = 4

Hence, central atom is $sp^3$ hybridized.

(iv) $\mathrm{H_2O}$

Central atom: $\mathrm{O}$
Oxygen forms 2 $\sigma$ -bonds with H atoms
Oxygen has 2 lone pairs

So,

\text{Steric number} = 2 + 2 = 4

Hence, central atom is $sp^3$ hybridized.

(v) $\mathrm{NO_2}$

Central atom: $\mathrm{N}$
Nitrogen forms 2 $\sigma$ -bonds with O atoms
One unpaired electron is present
Geometry is bent, electron domain arrangement is trigonal planar

Thus the central atom is generally taken as $sp^2$ hybridized.

Evaluate each option

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

$\mathrm{BF_3}$ : $sp^2$
$\mathrm{NO_2^-}$ : $sp^2$

✅ Both are $sp^2$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

$\mathrm{NH_2^-}$ : $sp^3$
$\mathrm{BF_3}$ : $sp^2$

❌ Not both $sp^2$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

$\mathrm{NH_2^-}$ : $sp^3$
$\mathrm{H_2O}$ : $sp^3$

❌ Not both $sp^2$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$

$\mathrm{H_2O}$ : $sp^3$
$\mathrm{NO_2}$ : $sp^2$

❌ Not both $sp^2$

Final answer

The correct pair is:

\boxed{\text{A: } \mathrm{BF_3} \text{ and } \mathrm{NO_2^-}}

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

So, they agree.

More from Chemical Bonding and Molecular Structure

View written solutionFree

Correct answer: A

Determine the hybridization of the central atom in each species

Hybridization depends on the steric number:

\text{Steric number} = \text{number of } \sigma\text{-bonds} + \text{number of lone pairs on central atom}

Steric number $= 3 \Rightarrow sp^2$
Steric number $= 4 \Rightarrow sp^3$

Check each molecule/ion

(i) $\mathrm{BF_3}$

Central atom: $\mathrm{B}$
Boron forms 3 $\sigma$ -bonds with 3 fluorine atoms
Lone pairs on B = 0

So,

\text{Steric number} = 3 + 0 = 3

Hence, central atom is $sp^2$ hybridized.

(ii) $\mathrm{NO_2^-}$

Central atom: $\mathrm{N}$
Nitrogen is bonded to two oxygens
Due to resonance, there are 2 $\sigma$ -bonds and 1 lone pair on N

So,

\text{Steric number} = 2 + 1 = 3

Hence, central atom is $sp^2$ hybridized.

(iii) $\mathrm{NH_2^-}$

Central atom: $\mathrm{N}$
Nitrogen forms 2 $\sigma$ -bonds with H atoms
Total valence electrons on N including negative charge give 2 lone pairs

So,

\text{Steric number} = 2 + 2 = 4

Hence, central atom is $sp^3$ hybridized.

(iv) $\mathrm{H_2O}$

Central atom: $\mathrm{O}$
Oxygen forms 2 $\sigma$ -bonds with H atoms
Oxygen has 2 lone pairs

So,

\text{Steric number} = 2 + 2 = 4

Hence, central atom is $sp^3$ hybridized.

(v) $\mathrm{NO_2}$

Central atom: $\mathrm{N}$
Nitrogen forms 2 $\sigma$ -bonds with O atoms
One unpaired electron is present
Geometry is bent, electron domain arrangement is trigonal planar

Thus the central atom is generally taken as $sp^2$ hybridized.

Evaluate each option

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

$\mathrm{BF_3}$ : $sp^2$
$\mathrm{NO_2^-}$ : $sp^2$

✅ Both are $sp^2$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

$\mathrm{NH_2^-}$ : $sp^3$
$\mathrm{BF_3}$ : $sp^2$

❌ Not both $sp^2$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

$\mathrm{NH_2^-}$ : $sp^3$
$\mathrm{H_2O}$ : $sp^3$

❌ Not both $sp^2$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$

$\mathrm{H_2O}$ : $sp^3$
$\mathrm{NO_2}$ : $sp^2$

❌ Not both $sp^2$

Final answer

The correct pair is:

\boxed{\text{A: } \mathrm{BF_3} \text{ and } \mathrm{NO_2^-}}

Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

So, they agree.

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

(i) $\mathrm{BF_3}$

(ii) $\mathrm{NO_2^-}$

(iii) $\mathrm{NH_2^-}$

(iv) $\mathrm{H_2O}$

(v) $\mathrm{NO_2}$

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$

More from Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure question

Chemical Bonding and Molecular Structure question

(i) $\mathrm{BF_3}$

(ii) $\mathrm{NO_2^-}$

(iii) $\mathrm{NH_2^-}$

(iv) $\mathrm{H_2O}$

(v) $\mathrm{NO_2}$

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$

More from Chemical Bonding and Molecular Structure

(i) BF3\mathrm{BF_3}BF3​

(ii) NO2−\mathrm{NO_2^-}NO2−​

(iii) NH2−\mathrm{NH_2^-}NH2−​

(iv) H2O\mathrm{H_2O}H2​O

(v) NO2\mathrm{NO_2}NO2​

Option A: BF3\mathrm{BF_3}BF3​ and NO2−\mathrm{NO_2^-}NO2−​

Option B: NH2−\mathrm{NH_2^-}NH2−​ and BF3\mathrm{BF_3}BF3​

Option C: NH2−\mathrm{NH_2^-}NH2−​ and H2O\mathrm{H_2O}H2​O

Option D: H2O\mathrm{H_2O}H2​O and NO2\mathrm{NO_2}NO2​

More from Chemical Bonding and Molecular Structure

(i) BF3\mathrm{BF_3}BF3​

(ii) NO2−\mathrm{NO_2^-}NO2−​

(iii) NH2−\mathrm{NH_2^-}NH2−​

(iv) H2O\mathrm{H_2O}H2​O

(v) NO2\mathrm{NO_2}NO2​

Option A: BF3\mathrm{BF_3}BF3​ and NO2−\mathrm{NO_2^-}NO2−​

Option B: NH2−\mathrm{NH_2^-}NH2−​ and BF3\mathrm{BF_3}BF3​

Option C: NH2−\mathrm{NH_2^-}NH2−​ and H2O\mathrm{H_2O}H2​O

Option D: H2O\mathrm{H_2O}H2​O and NO2\mathrm{NO_2}NO2​

More from Chemical Bonding and Molecular Structure

(i) $\mathrm{BF_3}$

(ii) $\mathrm{NO_2^-}$

(iii) $\mathrm{NH_2^-}$

(iv) $\mathrm{H_2O}$

(v) $\mathrm{NO_2}$

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$

(i) $\mathrm{BF_3}$

(ii) $\mathrm{NO_2^-}$

(iii) $\mathrm{NH_2^-}$

(iv) $\mathrm{H_2O}$

(v) $\mathrm{NO_2}$

Option A: $\mathrm{BF_3}$ and $\mathrm{NO_2^-}$

Option B: $\mathrm{NH_2^-}$ and $\mathrm{BF_3}$

Option C: $\mathrm{NH_2^-}$ and $\mathrm{H_2O}$

Option D: $\mathrm{H_2O}$ and $\mathrm{NO_2}$