Structure of Atom Test

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Structure of Atom Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

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

A

B

C

D

Solution

Valence unpaired electrons are in 3p

Valence unpaired electrons are in 3p.

Thus, (d).
Question 2
4 / -1

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

A
n=3, l=0

B
n=3, l=1

C
n=2, l=0

D
n=2, l=1

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
4 / -1

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

A

B

C

D

Solution
Question 4
4 / -1

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

A

B

C

D

Solution
Question 5
4 / -1

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

A
2 and 0

B
0 and 2

C
1 and 2

D
2 and 1

Solution

Radial nodes = (n - l - 1)
Angular nodes = l, total nodes = (n - 1), nodal plane = l
Question 6
4 / -1

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

A
1s

B
2p

C
3s

D
3d

Solution
Question 7
4 / -1

Following suborbits with values of n and l are given
Q. Increasing order of energy of these suborbits is

A
IV < III < II < I

B
IV < III < I < II

C
IV < II < III < I

D
I < III < II < IV

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
4 / -1

The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is
[JEE Main 2013]

A

B

C

D

Solution
Question 9
4 / -1

For Cr(24), (EC) : [Ar] 4s¹ 3d⁵ The number of electrons with l = 1 to l = 2 are respectively

A
12 and 4

B
12 and 5

C
16 and 4

D
16 and 5

Solution

EC of Cr (24) is
Question 10
4 / -1

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?

A

B

C

D

Solution

4l -orbital
Thus, n = 4, l = 3
m, (anyone) = - 3, - 2, - 1, 0, +1, + 2, + 3
Question 11
4 / -1

Which orbital gives an electron, a greater probability being found close to the nucleus?

A
3p

B
3d

C
3s

D
equal

Solution

3s is spherically symmetrical and its electron density is maximum at the nucleus. It decreases with r.
Question 12
4 / -1

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

A
a

B
b

C
c

D
d

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
Question 13
4 / -1

‘Hartree’ is the atomic unit of energy and is equal to

A
-13 .6 eV atom^-1

B
-27.2 eV atom ^-1

C
+13.6 eV atom ^-1

D
+27.2 eV atom^-1

Solution

(a) 2s (b) 2 p (c) 3d (d) 3d
Thus, (c) and (d), both 3d, have equal energy.
Question 14
4 / -1

Which of the following electronic transitions requires that the greatest quantity of energy be absorbed by a hydrogen atom ?

A
n = 1 to n = 2

B
n = 2 to n = 4

C
n = 3 to n = 6

D
n = 3 to n = ∞

Solution

Therefore, electronic transition (a) requires greatest quantity of energy.
Question 15
4 / -1

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

A
Principal quantum number

B
Azimuthal quantum number

C
Spin quantum number

D
Magnetic quantum number

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
4 / -1

Quantum Numbers are solutions of _____________

A
Heisenberg’s Uncertainty Principle

B
Einstein’s mass energy relation

C
Schrodinger’s Wave Equation

D
Hamiltonian Operator

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
4 / -1

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

A
m

B
n

C
l

D
s

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
4 / -1

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

A
a

B
b

C
c

D
d

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
4 / -1

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

A
O₂^2-

B
O₂

C
O₂⁺

D
O₂^-

Solution

MO configuration of O₂⁺ (8+8-1=15)
Question 20
4 / -1

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

A
9

B
8

C
4

D
2

Solution

(9) n = 3 can have suborbits

Total nine electrons.