Question

JEE MAINS CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

=> ratio of ( Ro) 1 and ( Ro) 2 = 3:2

=> option [ 3 ] is correct.

EXPLANATION.

$\sf \to \: difference \: in \: \: radius \: \: of \: \: 3rd \: and \: 4th \: orbit \\ \\ \sf \to \: formula \: of \: radius \: = 0.529 \times \dfrac{ {n}^{2} }{z} \\ \\ \sf \to \: difference \: of \: 3rd \: and \: 4th \: orbital \: of \: he {}^{ + } \\ \\ \sf \to \: 0.529 \times \frac{( {4}^{2} - {3}^{2} )}{2} ......(1)$

$\sf \to \: difference \: of \: 3rd \: and \: 4th \: orbital \: of \: li {}^{ + } \\ \\ \sf \to \: 0.529 \times \frac{( {4}^{2} - {3}^{2}) }{3} ......(2)$

$\sf \to \: according \: to \: the \: question \\ \\ \sf \to \: ratio \: of \: r_{o}(1) \: \: and \: \: r_{o}(2) \\ \\ \sf \to \: \dfrac{ \dfrac{0.529 \times ( {4}^{2} - {3}^{2} )}{2} }{ \dfrac{0.529 \times ( {4}^{2} - {3}^{2} }{3} } \\ \\ \\ \sf \to \: \frac{ \cancel{0.529 \times ( {4}^{2} - {3}^{2}) }}{2} \times \frac{3}{ \cancel{0.529 \times ( {4}^{2} - {3}^{2} )}} \\ \\ \sf \to \: \frac{ r_{o}(1) }{ r_{o}(2) } = \frac{3}{2}$

Answer 2

${ \huge \color{red} { \underline{ \rm \color{black}{To \: find }}}}$

The ratio of radius of 3rd and 4th orbital of H$\sf {e}^{ + }$ ions and L.ions.

${ \huge{ \underline{ \underline{ \rm{Given}}}}}$

Difference in the radius of 3rd and 4th orbital is $\sf(R_O)1 and (R_O)2$

$\huge \color{blue} \underline \color{black} \rm{Solution}$

$\sf \: Formula \: of \: radius=0.529 \times \frac{ {n}^{2} }{z}$

Difference in 3rd and 4th orbital of h$\sf {e}^{ + }$

$\sf \rightarrow0.529 \times \frac{ {4}^{2} - {3}^{2} }{2} ....(i)$

Difference in 3rd and 4th orbital of l$\sf {i}^{2 + }$

$\rightarrow \sf0.529 \times \frac{ {4}^{2} - {3}^{2} }{3} ..(ii)$

According to the question; the ratio of $\sf \: R_O(1) \: and \: R_O(2)$

$\large \rightarrow \sf \frac{ \frac{0.529 \times ( {4}^{2} - {3}^{2} )}{2} }{ \frac{0.529 \times ( {4}^{2} - {3}^{2} )}{3} }$

$\sf \implies \frac{ \cancel{0.529 }\times ( \cancel{{4}^{2}} - \cancel{{3}^{2}} ) }{2} \frac{3}{ \cancel{0.529} \times( \cancel{ {4}^{2}} - \cancel{ {3}^{2}} )}$

$\sf \: \frac{R_O(1)}{R_O(2)} = \frac{3}{2}$

So the correct answer is option (c)