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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Chemistry
Chapter
Structure of Atom

JEE Advanced 2017 · Paper 1 · Chemistry · Structure of Atom

The wave function, ψ _ n, l, m _ l is a mathematical function whose...

The wave function, ψn,l,ml{ψ}_{n, l, {m}_{l}} is a mathematical function whose value depends upon spherical polar coordinates (r,θ,ϕ)\left(r, \theta, ϕ\right) of the electron and characterized by the quantum numbers n,landmln, l and {m}_{l} . Here rr is distance from nucleus, θ\theta is colatitude and ϕϕ is azimuth. In the mathematical functions given in the Table, Z is atomic number and ao{a}_{o} is Bohr radius.
 

Column – 1Column – 2Column – 3
(I) Is orbital(i) ψn,l,ml∝(Zao)32e−(Zreo){ψ}_{n, l, {m}_{l}} ∝ \left(\dfrac{Z}{{a}_{o}}\right)\dfrac{3}{2}{e}^{- \left(\dfrac{Zr}{{e}_{o}}\right)}(P) The wave function, ψ _ n, l, m _ l is a mathematical function whose value depends upon spherical polar coordinates (r,
(II) 2s orbital(ii) One radial node(Q) Probability density at nucleus ∝1ao3∝ \dfrac{1}{{a}_{o}^{3}}
(III) 2pz2{p}_{z} orbital(iii) ψn,l,ml∝(Zao)52re−(Zr2ao)cosθ{ψ}_{n, l, {m}_{l}} ∝ \left(\dfrac{Z}{{a}_{o}}\right)\dfrac{5}{2}{re}^{- \left(\dfrac{Zr}{2{a}_{o}}\right)}cos \theta(R) Probability density is maximum at nucleus
(IV) 3dz23{d}_{z}^{2} orbital(iv) xyxy -plane is a nodal plane(S) Energy needed to excite an electron from n=2n = 2 state to n=4n = 4 state is 2732\dfrac{27}{32} times the energy needed to excite electron from n=2n = 2 state to n=6n = 6 state.
For hydrogen atom, what is the only CORRECT combination?

Options

  1. A(I)(i)(P)( I ) ( i ) ( P )
  2. B(II)(i)(Q)( II ) ( i ) ( Q )
  3. C(I)(iv)(R)( I ) ( iv ) ( R )
  4. D(I)(i)(S)( I ) ( i ) ( S )

Correct answer

D. (I)(i)(S)( I ) ( i ) ( S )

Step-by-step solution

We have to select only correct combination hence, the option (iv) is correct.

For 1s orbital : ψn,l,mα(Za0)32e−zra0{ψ}_{n, l, m}\alpha\left(\dfrac{Z}{{a}_{0}}\right)\dfrac{3}{2}e\dfrac{- zr}{{a}_{0}}

Energy needed to excite an electron : from n=2n = 2 to n=4n = 4

ΔE2−4=13.6Z2×316eV{\Delta E}_{2 - 4} = 13 .6 {Z}^{2} \times \dfrac{3}{16}eV

Energy needed to excite an electron : from n=2n = 2 to n=6n = 6

ΔE2−6=13.6Z2×836{\Delta E}_{2 - 6} = 13 .6 {Z}^{2} \times \dfrac{8}{36}

ΔE2−4=2732E2−6{\Delta E}_{2 - 4} = \dfrac{27}{32}{E}_{2 - 6} (hence, true)

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