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This is a Multiple Choice Questions as classified in NCERT Exemplar

55. Options (i) and (iii) are correct since nucleophiles are negatively charged or electron rich (as contains lone pair of electrons) species. 

Kindly go through the solution

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This is a Multiple Choice Questions as classified in NCERT Exemplar

53. Options (i) and (ii) are correct since two or more compounds with same molecular formula but different functional groups are known as functional isomers. In the given compounds- 

I. Aldehydic group. II. Ketonic group. III. Ketonic group. IV. Aldehydic group. So, II and III, I and IV are not functional isomers.

(I) $\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right]+\left[\begin{array}{ll}a\hfill & b\hfill \\ b\hfill & a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a+a\hfill & \hfill b+b\hfill \\ \hfill -b+b\hfill & \hfill a+a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2a\hfill & \hfill 2b\hfill \\ \hfill 0\hfill & \hfill 2a\hfill \end{array}\right]$

This is a Multiple Choice Questions as classified in NCERT Exemplar

52. Option (ii) II and III since in position isomerism, two or more compounds differ in the position of substituents, functional group, or multiple bonds but their molecular formula is same. In pentanone-2 and pentanone-3, position of ketonic group is different. CH₃-CH₂-CH₂-CO-CH₃ and CH₃ - CH₂-CO-CH₂- CH₃

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A $=\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right],$ B $=\left[\begin{array}{rr}\hfill 1& \hfill 3\\ \hfill -2& \hfill 5\end{array}\right],$ C $=\left[\begin{array}{rr}\hfill -2& \hfill 5\\ \hfill 3& \hfill 4\end{array}\right]$

(i) A + B = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2+1\hfill & \hfill 4+3\hfill \\ \hfill 3-2\hfill & \hfill 2+5\hfill \end{array}\right]=\left[\begin{array}{ll}3\hfill & 7\hfill \\ 1\hfill & 7\hfill \end{array}\right]$

(ii) A - B = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-1\hfill & \hfill 4-3\hfill \\ \hfill 3-(-2)\hfill & \hfill 2-5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -3\hfill \end{array}\right].$

(iii) 3A - C = 3 $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{ll}3\times 2\hfill & 3\times 4\hfill \\ 3\times 3\hfill & 3\times 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill 12\hfill \\ \hfill 9\hfill & \hfill 6\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

$$ $=\left[\begin{array}{cc}\hfill 6-(-2)\hfill & \hfill 12-5\hfill \\ \hfill 9-3\hfill & \hfill 6-4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 8\hfill & \hfill 7\hfill \\ \hfill 6\hfill & \hfill 2\hfill \end{array}\right]$

(iv) AB = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ll}2\times 1+4\times (-2)\hfill & 2\times 3+4\times 5\hfill \\ 3\times 1+2\times (-2)\hfill & 3\times 3+2\times 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-8\hfill & \hfill 6+20\hfill \\ \hfill 3-4\hfill & \hfill 9+10\hfill \end{array}\right]$

= $\left[\begin{array}{rr}\hfill -6& \hfill 26\\ \hfill -1& \hfill 19\end{array}\right]$

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This is a Multiple Choice Questions as classified in NCERT Exemplar

51. Options (ii) and (iii) are correct since the electrophiles are positively charged or electron deficient species. They act as Lewis acids. AlCl₃, SO₃are Lewis acids while  NO₂⁺, CH₃⁺, CH₃-C⁺ = O are positively charged species. Therefore, the species involved in the above-mentioned options act as electrophiles.

A 3 × 3 order matrix will have 9 elements

(M) Since, the elements can be either 1 or $\mathrm{0il},$ number of choices for each element is 2 .

The required no. of arrangement = 29 (4,2 × 2 × 2 9 times ) $\Rightarrow$ 512

So, option D is correct.

This is a Multiple Choice Questions as classified in NCERT Exemplar

50. Options (i), (iii) and (iv) are correct since the spatial arrangement of groups or atoms can be checked by doing two interchanges and bringing the H atom below the plane of the paper. Then find out the sequence of the left-out groups in a particular order whether clockwise or anticlockwise starting from the atom with the highest atomic number to the atom with lower atomic numbers. So, out of the given options option (ii) has the same spatial arrangement as (A) while in rest three spatial arrangements are different.

This is a Multiple Choice Questions as classified in NCERT Exemplar

49. Options (i) and (iv) are correct since only in these two compounds all the carbon atoms are having same hybridization i.e., sp hybridized in option (i) and sp2 hybridized in option (iv).

For the matrices to be equal, the corresponding elements

(B) heed to be equal so,

Fora₁₁, 3x+ 7 = 0

$\Rightarrow$ 3x = -7.

$\Rightarrow$ x = $-\frac{7}{3}$

fora₁₂, 5 = y - 2 

$\Rightarrow$ y = + 5 + 2 = 7.

For a₂₁, y + 1 = 8

$$ $\Rightarrow$ y = 8 - 1 = 7.

fora₂₂, 2 - 3x = 4.

$\Rightarrow$ 3x = 2 - 4

$\Rightarrow x=-\frac{2}{3}$

As the variable x and y has more than one value which is not peacetable.

Option B is correct.

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