1

### JEE Main 2019 (Online) 9th January Morning Slot

The highest value of the calculated spin- only magnetic moment (in BM) among all the transition metal complexes is :
A
5.92
B
6.93
C
3.87
D
4.90

## Explanation

In transition metal contains d orbital, and in d orbital maximum no of unpaired electron possible = 5.

Spin only magnetic moment,

${\mu _{spin}} = \sqrt {n\left( {n + 2} \right)}$    B. M

here n $=$ Number of unpaired electrons.

$\therefore$    ${\mu _{spin}} = \sqrt {5\left( {5 + 2} \right)}$0    B. M

$=$ 5.92   B. M
2

### JEE Main 2019 (Online) 9th January Morning Slot

For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $\left( {\overline v } \right)$ against $\left( {{1 \over {{n^2}}}} \right)$ will be (The Rydberg constant, RH is in wave number unit)
A
Linear with intercept $-$RH
B
Non linear
C
Linear with slope RH
D
Linear with slope $-$RH

## Explanation

We know,

$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$

here ni $=$ 8 and nf $=$ n

Z $=$ 1 for hydrogen

$\therefore$   $\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$

So, graph is $\to$ So, graph will be linear with slope RH.
3

### JEE Main 2019 (Online) 9th January Evening Slot

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

(c) According to wave mechanics, the ground state angular momentum is equal to ${h \over {2\pi }}$.

(d) The plot of $\psi$ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.
A
(a), (b)
B
(a), (d)
C
(b), (c)
D
(a), (c)
4

### JEE Main 2019 (Online) 10th January Evening Slot

The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He+ ion in eV is :
A
$-$ 6.04
B
$-$ 54.4
C
$-$ 27.2
D
$-$ 3.4

## Explanation

(E)nth = (EGND)H . ${{{Z^2}} \over {{n^2}}}$

E3rd (He+) = ($-$13.6eV) . ${{{2^2}} \over {{3^2}}}$ = $-$ 6.04 eV