Chemical Bonding and Molecular Structure — NEET UG practice

1. The formal charge on an atom in a Lewis structure is calculated using the formula:

$$\text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons})$$

Here, oxygen has 6 valence electrons.

2. For oxygen atom 2, it has 4 bonding electrons (two shared bonds) and 4 non-bonding electrons, so:

$$\text{Formal charge} = 6 - 4 - \frac{1}{2}(4) = 0$$

3. For oxygen atom 1, it has 6 bonding electrons and 2 non-bonding electrons, so:

$$\text{Formal charge} = 6 - 2 - \frac{1}{2}(6) = +1$$

4. For oxygen atom 3, it has 2 bonding electrons and 6 non-bonding electrons, so:

$$\text{Formal charge} = 6 - 6 - \frac{1}{2}(2) = -1$$

5. So the formal charges on oxygen atoms 2, 1 and 3 respectively are:

$$0,\ +1,\ -1$$

Hence, the answer is B.

1. To match each species with its geometry, first work out the hybridisation of the central atom using the number of bonding pairs and lone pairs around it.

$\mathrm{PCl}_5$ has phosphorus surrounded by 5 bond pairs and no lone pair, so it is $sp^3d$ hybridised, giving a trigonal bipyramidal shape (III).

2. For $\mathrm{BrF}_5$, bromine has 5 bonding pairs and 1 lone pair, which needs $sp^3d^2$ hybridisation. With one octahedral position occupied by the lone pair, the resulting geometry is:

$$sp^3d^2 \rightarrow \text{square pyramidal (IV)}$$

3. In $\mathrm{BF}_4^{-}$, boron forms 4 bond pairs with no lone pair, so it is $sp^3$ hybridised, giving a tetrahedral shape (I).

4. In $[\mathrm{Ni(CN)}_4]^{2-}$, nickel uses $dsp^2$ hybridisation (4 bond pairs, no lone pair, involving a d-orbital due to the strong-field $\mathrm{CN}^-$ ligands), giving a square planar geometry (II).

5. Putting these together:

$$A-III,\ B-IV,\ C-I,\ D-II$$

Hence, the answer is C.

1. To match each molecule, count the number of sigma ($\sigma$) bonds, pi ($\pi$) bonds, and lone pairs present in its structure.

$\mathrm{C}_2\mathrm{H}_4$ (ethylene) has one $\mathrm{C=C}$ double bond and four $\mathrm{C-H}$ bonds:

A double bond consists of one $\sigma$ bond and one $\pi$ bond, so counting all bonds gives:

$$\sigma = 5,\quad \pi = 1$$

This matches IV.

2. $\mathrm{C}_2\mathrm{H}_2$ (acetylene) is written as $\mathrm{H-C \equiv C-H}$, where the triple bond contains one $\sigma$ bond and two $\pi$ bonds:

$$\sigma = 3,\quad \pi = 2$$

This matches I.

3. $\mathrm{CH}_4$ has carbon forming four single $\mathrm{C-H}$ bonds, all of which are $\sigma$ bonds:

$$\sigma = 4$$

This matches III.

4. $\mathrm{NH}_3$ has nitrogen forming three $\mathrm{N-H}$ sigma bonds while keeping one lone pair:

$$\sigma = 3,\ \text{one lone pair}$$

This matches II.

5. Combining all matches:

$$A-IV,\ B-I,\ C-III,\ D-II$$

Hence, the answer is B.