Chemistry - CHEMICAL BONDING Question with Solution | TestHub

To determine the number of lone pairs on the central atom, we use the VSEPR theory. The steric number (SN) is calculated as the sum of sigma bonds and lone pairs.

1. ClF₄⁺: Central atom Cl. Valence electrons = 7. Charge = +1. Number of F atoms = 4.

Number of valence electrons available for bonding and lone pairs = $7-1=6$.

Number of sigma bonds = 4.

Number of lone pairs = $(6-4)/2=1$. (NOT 2 lone pairs)

2. SeF₂: Central atom Se. Valence electrons = 6. Number of F atoms = 2.

Number of valence electrons available for bonding and lone pairs = 6.

Number of sigma bonds = 2.

Number of lone pairs = $(6-2)/2=2$. (2 lone pairs)

3. XeO₃: Central atom Xe. Valence electrons = 8. Number of O atoms = 3.

Number of valence electrons available for bonding and lone pairs = 8.

Number of sigma bonds = 3 (each double bond to oxygen counts as one sigma bond for VSEPR).

Number of lone pairs = $(8-3\times 2)/2=(8-6)/2=1$. (NOT 2 lone pairs)

(Alternatively, if considering all valence electrons: $8-(3\times 2)=2$ electrons remaining, so 1 lone pair.)

4. ClF₂⁺: Central atom Cl. Valence electrons = 7. Charge = +1. Number of F atoms = 2.

Number of valence electrons available for bonding and lone pairs = $7-1=6$.

Number of sigma bonds = 2.

Number of lone pairs = $(6-2)/2=2$. (2 lone pairs)

5. SNF₃: Central atom S. Valence electrons = 6. Number of N atoms = 1, F atoms = 3.

Number of valence electrons available for bonding and lone pairs = 6.

Number of sigma bonds = 4 (S-N, 3 S-F).

Number of lone pairs = $(6-4)/2=1$. (NOT 2 lone pairs)

6. I₃⁻: Central atom I. Valence electrons = 7. Charge = -1. Number of terminal I atoms = 2.

Number of valence electrons available for bonding and lone pairs = $7+1=8$.

Number of sigma bonds = 2.

Number of lone pairs = $(8-2)/2=3$. (NOT 2 lone pairs)

7. SOCl₂: Central atom S. Valence electrons = 6. Number of O atoms = 1, Cl atoms = 2.

Number of valence electrons available for bonding and lone pairs = 6.

Number of sigma bonds = 3 (S-O, S-Cl, S-Cl).

Number of lone pairs = $(6-3\times 2)/2=(6-6)/2=0$. (NOT 2 lone pairs)

(Alternatively, if considering all valence electrons: $6-(1\times 2+2\times 1)=2$ electrons remaining, so 1 lone pair. The solution's calculation of 1 lone pair is correct.)

8. SO₃²⁻: Central atom S. Valence electrons = 6. Charge = -2. Number of O atoms = 3.

Number of valence electrons available for bonding and lone pairs = $6+2=8$.

Number of sigma bonds = 3.

Number of lone pairs = $(8-3\times 2)/2=(8-6)/2=1$. (NOT 2 lone pairs)

Compounds with exactly two lone pairs on the central atom are SeF₂ and ClF₂⁺.

So, the total number of such compounds, $X=2$.

The question asks for $\frac{X}{2}$.

$\frac{X}{2}=\frac{2}{2}=1$.

The final answer is $\boxed{1}$.