Select the CORRECT statement. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. Acidic strength increases down the group as bond dissociation energy decreases,so $H_2S > H_2O$. Thus,option $A$ is incorrect.$2$. The ${p_\p

Vedclass · Accepted Answer

Bond angle order: $OF_2 < H_2O < Cl_2O < ClO_2$

Vedclass · Answer

(D) $1$. Acidic strength increases down the group as bond dissociation energy decreases,so $H_2S > H_2O$. Thus,option $A$ is incorrect.$2$. The ${p_\pi} - {p_\pi}$ bonding molecular orbital (bonding $M.O.$) formed by the overlap of $p$-orbitals has $\psi$ (gerade) symmetry,while the antibonding $M.O.$ has $\psi$ (ungerade) symmetry. However,the statement in $B$ is often debated based on specific orbital orientation; let's evaluate others.$3$. $NCl_3$ hydrolyzes to form $NH_3$ and $HOCl$ because $Cl$ is more electronegative than $N$. $NF_3$ is resistant to hydrolysis because $N$ is more electronegative than $F$ and there is no vacant $d$-orbital on $N$. Thus,option $C$ is incorrect.$4$. Bond angles: $OF_2$ $(103.2^\circ)$,$H_2O$ $(104.5^\circ)$,$Cl_2O$ $(110.9^\circ)$,$ClO_2$ $(117.4^\circ)$. The order $OF_2 < H_2O < Cl_2O < ClO_2$ is correct due to increasing steric repulsion and electronegativity effects. Therefore,option $D$ is the correct statement.

List-$I$ (Molecule / Species)	List-$II$
$(P)$ $NO_2$ (Unpaired electron in)	$(1)$ Vacant $p$-orbital involved in hybridization
$(Q)$ $B_2H_6$	$(2)$ $sp^2$-orbital
$(R)$ $ClO_3$ (Unpaired electron in)	$(3)$ $sp^3$-orbital
$(S)$ $CH_3^+$	$(4)$ $120^\circ$ Bond angle

Select the $CORRECT$ statement.

Similar Questions

What is the formal charge on the $C$ atom in the structure $S \equiv C - \ddot{N} :$?

Which of the following does not undergo Lewis acid-base reaction?

$S-O$ bond length is maximum in

Which of the following molecules do not exist?

Vedclass Test Series

Exam Paper Generator

Online Exam Module