Class 12th

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iv)

The E-H bond dissociation energy of SbH₃ is the lowest among the aforementioned compounds, it easily releases H, making it a strong reducing agent. The greater the lowering agent, the weaker the E-H bond.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (i)

Fluorine is the most reactive element, forming hydrogen fluoride when it combines with hydrogen gas.

Due to the shorter bond length between the two atoms, it takes a lot of energy to break the link between hydrogen and fluorine.

As a result, as we move down the group, the bond dissociation energy drops.

This is a  Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}?P\left(A/B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ \Rightarrow P\left(A\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}\\ \Rightarrow P\left(A\cap B\right)=P\left(A\right).P\left(B\right)\\ So,AisindependentofB.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (i)

The compound has the following number of electrons:

NO₃^-  =  32e^- 

CO₃^2-  =  32e^- 

ClO₃^-  =  42e^- 

SO₃ ^-  =  42e^- 

Hence,  NO₃^- and CO₃^2-  are isoelectronic

This is a  Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2\\ ={{\displaystyle \sum _{}^{}{X}^{2}P\left(X\right)-\left[{\displaystyle \sum _{}^{}X.P\left(X\right)}\right]}}^2={\displaystyle \sum _{}^{}{p}_i{x}_i^2}-{\left({\displaystyle \sum _{}^{}{p}_i{x}_i}\right)}^2\\ Hence,Var\left(X\right)isequalto{\displaystyle \sum _{}^{}{p}_i{x}_i^2}-{\left({\displaystyle \sum _{}^{}{p}_i{x}_i}\right)}^2.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (iii)

If an electron pair is donated to a vacant orbital on one atom and a lone pair of electrons on the other, the bonding is designated pπ-dπ depending on the orbital to which the electron pair is donated and the orbital from which the electron pair is donated.

Because their valence shells lack d-orbitals, nitrogen, carbon, and boron cannot form a pi bond. Phosphorus, on the other hand, can produce pi bonds.

This is a  Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

This is a  Fill in the Blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,P\left(A\text{'}\cup B\text{'}\right)=\frac{2}{3}andP\left(A\cup B\right)=\frac{5}{9}\\ \therefore 1-P\left(A\cap B\right)=\frac{2}{3}\\ \Rightarrow P\left(A\cap B\right)=1-\frac{2}{3}=\frac{1}{3}\\ NowP\left(A\text{'}\right)+P\left(B\text{'}\right)=1-P\left(A\right)+1-P\left(B\right)=2-\left[P\left(A\right)+P\left(B\right)\right]\\ =2-\left[P\left(A\cup B\right)+P\left(A\cap B\right)\right]\\ =2-\left[\frac{5}{9}+\frac{1}{3}\right]=2-\frac{8}{9}=\frac{10}{9}\\ Hence,thevalueofthefilleris\frac{10}{9}.\end{array}$