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QUESTION: 1

**Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. Which is a possible set of quantum numbers for a valence unpaired electrons in ground state atom of phosphorus (Z = 15)?**

Solution:

Valence unpaired electrons are in 3p

Valence unpaired electrons are in 3p.

Thus, (d).

QUESTION: 2

For a multi-electron atom, set of quantum numbers is given as

2,0,0,1/2 ; 2,0,0,-1/2

**Q. Thus, the next higher allowed set of n and / quantum numbers for this atom in its ground state is**

Solution:

Given a set of quantum numbers, n=2,l=0 for a multi-electron atom refers to 2s orbital.

The next higher allowed set of 'n' and 'l' quantum numbers for this atom in the ground state is n=2,l=1. This corresponds to 2p orbital.

Note: The orbital with a higher value of the sum (n+l) has higher energy.

For 2s orbital (n+l)=(2+0)=2

For 2p orbital (n+l)=(2+1)=3

QUESTION: 3

Magnetic moments of the following isoelectronic species (24 electrons) are in order

Solution:

QUESTION: 4

The orbital angular momentum for a d-orbital electron is given by

Solution:

QUESTION: 5

The number of radial nodes in 3s and 2p respectively are

Solution:

Radial nodes = (n - l - 1)

Angular nodes = l, total nodes = (n - 1), nodal plane = l

QUESTION: 6

A hydrogen like species in fourth orbit has radius 1.5 times that of Bohr's orbit. In neutral state, its valence electron is in

Solution:

QUESTION: 7

Following suborbits with values of n and l are given

**Q. Increasing order of energy of these suborbits is**

Solution:

By Aufbau rule, smaller the value of (n + I), smaller the energy. If (n + I) is same for two or more orbits, suborbit with lower value of n, has smaller energy.

QUESTION: 8

The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is

**[JEE Main 2013]**

Solution:

QUESTION: 9

For Cr(24), (EC) : [Ar] 4s^{1} 3d^{5} The number of electrons with l = 1 to l = 2 are respectively

Solution:

EC of Cr (24) is

QUESTION: 10

Last filling electron in lanthanides find place in 4f-orbital. Which of the following sets of quantum number is correct for an electron in 4f orbital?

Solution:

4l -orbital

Thus, n = 4, l = 3

m, (anyone) = - 3, - 2, - 1, 0, +1, + 2, + 3

*Multiple options can be correct

QUESTION: 11

**Direction (Q. Nos. 11-15) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.**

**Q. Quantum numbers have been matched with their characteristics. Select correct matching(s).**

Solution:

(a) Three-dimensional shape of orbital is derived from angular quantum number: True

(d) Spin of the electron is derived from spin quantum number: True

QUESTION: 12

In the following pair, each has two orbitals I and II. Select the ones in which II experiences larger effective nuclear charge than l

Solution:

**Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electrons in it.**

**Since 3p is closer to nucleus than 3d so it will experience greater Z**

**Hence C**

*Multiple options can be correct

QUESTION: 13

In a multi-electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields?

Solution:

(a) 2s (b) 2 p (c) 3d (d) 3d

Thus, (c) and (d), both 3d, have equal energy.

*Multiple options can be correct

QUESTION: 14

Which of the following statements is/are correct?

Solution:

(a) Cr (24) = [Ar] 4s^{1}3d^{5} Correct

(b) K = 1s^{2}2s^{2}2 p^{6}3s^{2}3p^{6}4s^{1}

(c) p-suborbit has three orbitals, each orbital can have maximum of two electrons thus, correct.

(d) s-suborbit has one orbital and this orbital can have maximum of two electrons thus, correct.

QUESTION: 15

The quantum number which specifies the location of an electron as well as energy is

Solution:

According to the given sets of options, only option d is correct. The reason or exceptional configuration of Cr is stability of half-filled d subshell. If the Aufbau principle was not followed then the configuration of Cr be [Ar] 3d6. So, if the Aufbau principle won’t be followed, then the options won’t have an electron in their 4s subshell.

QUESTION: 16

Quantum Numbers are solutions of _____________

Solution:

When the wave function for an atom is solved using the Schrodinger Wave Equation, the solutions obtained are called the Quantum Number which are basically n, l and m.

QUESTION: 17

Which quantum numbers gives the shell to which the electron belongs?

Solution:

The principal quantum number, n, gives the shell to which the electron belongs. The energy of the shell is dependent on ‘n’.

QUESTION: 18

**Direction (Q. Nos. 18) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.**

**Q. Match the entries in Column I with correctly related quantum number(s) in Column II **

Solution:

(i) Orbital angular momentum L =

L depends on the value of l (azimuthal quantum number)

(ii) To describe wave function (), n, I and m are needed, if, it obeys Pauli's exclusion principle, then s is also needed.

(iii) Value of n, I and m are needed to determine size, shape and orientation.

(iv) Probability density (^{2}) is based on n, I and m

QUESTION: 19

**Direction (Q. Nos. 19 and 20) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**

**Q.** Bond order of 1.5 is shown by

Solution:

MO configuration of O_{2}^{+} (8+8-1=15)

*Answer can only contain numeric values

QUESTION: 20

The maximum number of electrons that can have principal quantum number, n = 3 and spin quantum number,

Solution:

(9) n = 3 can have suborbits

Total nine electrons.

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