VSEPR Theory Questions in English

(A) In $XeO_6^{4-}$ ion, the central atom is Xenon $(Xe)$.
The oxidation state of $Xe$ is calculated as: $x + 6(-2) = -4$, which gives $x = +8$.
$Xe$ has $8$ valence electrons. In the $XeO_6^{4-}$ ion, $Xe$ forms $6$ double bonds with $6$ oxygen atoms.
Each double bond consists of $2$ electron pairs (one sigma and one pi bond), but in Lewis structure counting for geometry/$VSEPR$, we count the number of electron domains around the central atom.
Here, $Xe$ is bonded to $6$ oxygen atoms, so there are $6$ bonding domains (bond pairs).
Since all $8$ valence electrons of $Xe$ are used in bonding with $6$ oxygen atoms (each oxygen atom requires $2$ electrons to complete its octet, and the negative charge accounts for the extra electrons), there are no lone pairs remaining on the $Xe$ atom.
Total number of electron pairs around the central atom = $6 \text{ (bond pairs)} + 0 \text{ (lone pairs)} = 6$.

(B) In $\text{ClF}_3$, the central chlorine atom has $7$ valence electrons.
It forms $3$ covalent bonds with fluorine atoms, leaving $2$ lone pairs on the chlorine atom.
According to the $VSEPR$ theory, the steric number is $3 + 2 = 5$, which corresponds to $sp^3d$ hybridization.
The presence of $2$ lone pairs in the equatorial positions of the trigonal bipyramidal electron geometry results in a $T$-shaped molecular geometry.

(D) Step $1$: Calculate the number of valence electrons in $IF_5$: $7 + 5 \times 7 = 42$ electrons.
Step $2$: Determine the hybridization: The central atom $I$ is bonded to $5$ $F$ atoms and has $1$ lone pair, giving a steric number of $6$ ($sp^3d^2$ hybridization).
Step $3$: Predict the geometry: With $5$ bond pairs and $1$ lone pair, the molecular geometry is square pyramidal.

(C) The central atom $Br$ has $7$ valence electrons.
It forms $3$ covalent bonds with $3$ $F$ atoms, utilizing $3$ electrons.
Remaining electrons = $7 - 3 = 4$ electrons.
These $4$ electrons form $2$ lone pairs.
Thus, there are $2$ lone pairs and $3$ bond pairs.

(B) According to the $VSEPR$ (Valence Shell Electron Pair Repulsion) theory, the magnitude of repulsion between electron pairs follows the order: $Lp-Lp > Lp-Bp > Bp-Bp$.
This is because a lone pair $(Lp)$ occupies more space around the central atom than a bond pair $(Bp)$, leading to greater repulsion.

(B) $1$. Chalcogens (Group $16$ elements) have $6$ electrons in their valence shell.
$2$. In their hydrides $(H_2X)$, the central chalcogen atom forms $2$ covalent bonds with hydrogen atoms.
$3$. Out of $6$ valence electrons, $2$ are used in bonding, leaving $4$ electrons as non-bonding electrons.
$4$. These $4$ non-bonding electrons form $2$ lone pairs.
$5$. Therefore, there are $2$ bond pairs and $2$ lone pairs.

(C) $1$. Determine the hybridization and geometry of each molecule:
$2$. $SO_2$: $sp^2$ hybridized, bent geometry.
$3$. $BCl_3$: $sp^2$ hybridized, trigonal planar geometry.
$4$. $CO_2$: $sp$ hybridized, linear geometry with a bond angle of $180^\circ$.
$5$. $NH_3$: $sp^3$ hybridized, trigonal pyramidal geometry.
$6$. Therefore, $CO_2$ is the linear molecule.

(D) The formal charge is calculated using the formula: $\text{Formal Charge} = V - L - \frac{1}{2}B$, where $V$ is the number of valence electrons, $L$ is the number of lone pair electrons, and $B$ is the number of bonding electrons.
For the sulphur atom in $H_2SO_4$:
$1$. Valence electrons $(V)$ = $6$.
$2$. Lone pair electrons $(L)$ = $0$ (all valence electrons are involved in bonding).
$3$. Bonding electrons $(B)$ = $12$ (sulphur forms $6$ bonds: $2$ double bonds with oxygen atoms and $2$ single bonds with hydroxyl groups).
$4$. $\text{Formal Charge} = 6 - 0 - \frac{1}{2}(12) = 6 - 6 = 0$.