Class 12th

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,n=10,p=\frac{1}{2},q=1-\frac{1}{2}=\frac{1}{2}\left(fortrue/falsequestions\right)\\ andr\ge 8i.e.,8,9,10\\ \therefore P\left(X\ge 8\right)=P\left(x=8\right)+P\left(x=9\right)+P\left(x=10\right)\\ ={}_{10}{C}_8{\left(\frac{1}{2}\right)}^8{\left(\frac{1}{2}\right)}^2+{}_{10}{C}_9{\left(\frac{1}{2}\right)}^9\left(\frac{1}{2}\right)+{}_{10}{C}_{10}{\left(\frac{1}{2}\right)}^{10}{\left(\frac{1}{2}\right)}^0\\ =45.{\left(\frac{1}{2}\right)}^{10}+10.{\left(\frac{1}{2}\right)}^{10}+{\left(\frac{1}{2}\right)}^{10}={\left(\frac{1}{2}\right)}^{10}\left(45+10+1\right)\\ =56*\frac{1}{1024}=\frac{7}{128}\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}ProbabilityofgettingQueen=\frac{4}{52}\\ So,therequiredprobability=P\left(Queen\right).P\left(Queen\right)\\ \text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}\text{?}=\frac{4}{52}.\frac{4}{52}=\frac{1}{13}.\frac{1}{13}\left(withreplacement\right)\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}WeknowthatforaBinomialdistribution,theoutcomesmustnotbedependentoneachother.\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

Oxygen is more electronegative than chlorine, the negative charge on chlorine disperses from ClO⁻ to ClO₄⁻ ion as the number of oxygen atoms linked to chlorine grows. As a result, ion stability will improve in the following order:

As the stability of the conjugate base rises, the acidic strength of the corresponding acid increases in the following order:

HClO < HClO₂ < HClO₃ < HClO₄

This is a short answer type question as classified in NCERT Exemplar

NO₃⁻ + 3Fe₂⁺ + 4H⁺→ NO⁺ + 3Fe₃⁺ + 2H₂O

[Fe (H₂O)₆]²⁺ + NO → [Fe (H₂O)₅ (NO)]²⁺ + H₂O ( brown ring )

This is a short answer type question as classified in NCERT Exemplar

A is PCl₅ (yellowish white powder)

P₄ + 10Cl₂→4PCl₅

B is PCl₃ (colourless oily liquid)

Hydrolysis of the following product

PCl₃+3H₂O→H3PO₃+3HCl

PCl₅+4H₂O→H3PO₄+5HCl

This is a short answer type question as classified in NCERT Exemplar

Six F atoms can be accommodated around sulphur due to fluorine's tiny size, however the chloride ion is comparably bigger, resulting in interatomic repulsion.

This is a short answer type question as classified in NCERT Exemplar

Oxygen is more electronegative than sulphur, the bond angle of H₂O is greater, and the bod pair electron of an OH bond will be closer to oxygen, causing bond-pair repulsion between bond pairs of two OH bonds

This is a short answer type question as classified in NCERT Exemplar

Chlorine has vacant d-orbitals, which become excited upon bonding when electrons from the 3p-orbital are promoted to the 3d-orbital, giving it a covalency of three.

Due to the lack of unoccupied d-orbitals in the second energy shell, fluorine cannot expand its octet. As a result, it can only have one covalency.