Chemistry Q134 – AP-EAMCET 2026

Home Chemistry Chemical BondingQuestion #208

B C l_{3}

on hydrolysis gives a complex ion X.

A l C l_{3}

in acidified aqueous solution forms a complex ion Y. The hybridization of central atoms in X and Y are respectively

Clue 1 of 3: Identify the complexes formed

First, consider the hydrolysis product of $B C l_{3}$ . Then, consider the form $A l C l_{3}$ takes in an acidified aqueous solution. These are your complex ions X and Y.

🧠 Full Solution Path

Step 1: Determine the complex ion X from

B C l_{3}

hydrolysis✦ Active

When $B C l_{3}$ undergoes hydrolysis, it forms boric acid, $B (O H)_{3}$ . Boric acid is a weak Lewis acid and in an aqueous solution, it accepts a hydroxyl ion from water to form a complex ion.

B C l_{3} + 3 H_{2} O \to B (O H)_{3} + 3 H C l

The complex ion X is formed when $B (O H)_{3}$ acts as a Lewis acid:

B (O H)_{3} + H_{2} O ⇌ [B (O H)_{4}]^{-} + H^{+}

So, complex ion X is $[B (O H)_{4}]^{-}$ . The central atom is Boron (B).

💡 Teacher's Secret Hint

Remember that boric acid, $B (O H)_{3}$ , does not directly release $H^{+}$ but rather accepts $O H^{-}$ from water, making it a Lewis acid.

Step 2: Determine the hybridization of the central atom in X○ Expand

In $[B (O H)_{4}]^{-}$ , the central Boron atom is bonded to four oxygen atoms (from four $O H$ groups). There are no lone pairs on the Boron atom. Therefore, the steric number for Boron is 4 (4 bond pairs + 0 lone pairs).

A steric number of 4 corresponds to $s p^{3}$ hybridization.

💡 Teacher's Secret Hint

To determine hybridization, count the number of sigma bonds and lone pairs around the central atom. This sum gives the steric number.

Step 3: Determine the complex ion Y from

A l C l_{3}

in acidified aqueous solution○ Expand

When $A l C l_{3}$ is dissolved in an acidified aqueous solution, the $A l^{3 +}$ ion gets highly hydrated to form a stable coordination complex. The most common hydrated aluminum ion is hexahydrate.

A l C l_{3} + 6 H_{2} O \to [A l (H_{2} O)_{6}]^{3 +} + 3 C l^{-}

So, complex ion Y is $[A l (H_{2} O)_{6}]^{3 +}$ . The central atom is Aluminum (Al).

💡 Teacher's Secret Hint

Aluminum, being a d-block element in this context, can expand its octet. Common coordination numbers for $A l^{3 +}$ in aqueous solutions are 6.

Step 4: Determine the hybridization of the central atom in Y○ Expand

In $[A l (H_{2} O)_{6}]^{3 +}$ , the central Aluminum atom is bonded to six oxygen atoms (from six water molecules). There are no lone pairs on the Aluminum atom. Therefore, the steric number for Aluminum is 6 (6 bond pairs + 0 lone pairs).

A steric number of 6 corresponds to $s p^{3} d^{2}$ hybridization.

💡 Teacher's Secret Hint

For elements in period 3 and beyond, d-orbitals become available for bonding, allowing for expanded octets and higher coordination numbers.

Step 5: Combine the hybridizations○ Expand

The hybridization of the central atom in X ( $[B (O H)_{4}]^{-}$ ) is $s p^{3}$ . The hybridization of the central atom in Y ( $[A l (H_{2} O)_{6}]^{3 +}$ ) is $s p^{3} d^{2}$ . Thus, the hybridizations are $s p^{3}, s p^{3} d^{2}$ respectively.

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Chemistry Question 134 – AP-EAMCET 2026

Clue 1 of 3: Identify the complexes formed

🧠 Full Solution Path