Practice Start Test

Structure of Atom

NEET Chemistry · 92 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
The number of d-electrons in Fe 2+ (Z = 26) is not equal to the number of electrons in which one of the following ?
A d-electrons in Fe (Z = 26)
B p-electrons in Ne (Z = 10)
C s-electrons in Mg (Z = 12)
D p-electrons in Cl(Z = 17)
Correct Answer
Option D
Solution

Number of d-electrons in Fe 2+ = 6 Number of p-electrons in Cl = 11

Q22
The angular momentum of electron in 'd' orbital is equal to
A 23h2\sqrt 3 h
B 0 ℏ\hbar
C 6 ℏ\sqrt 6 \,\hbar
D 2 ℏ\sqrt 2 \,\hbar
Correct Answer
Option C
Solution

We know Anglular momentum =

l(l+1)h\sqrt {l(l + 1)}h

For d obtial l = 2 ∴\therefore Angular momentum =

6h\sqrt {6}h
Q23
Calculate the energy in joule corresponding to light of wavelength 45 nm. (Planck,s constant, h = 6.63 × \times 10 −-34 J s, speed of light, c = 3 × \times 10 8 m s −-1 )
A 6.67 × \times 10 15
B 6.67 × \times 10 11
C 4.42 × \times 10 −-15
D 4.42 × \times 10 −-18
Correct Answer
Option D
Solution

E =

hcλhc \over \lambda

E =

6.63×10−34×3×10845×10−9{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {45 \times {{10}^{ - 9}}}}

= 4.42 ×\times 10 -18 J

Q24
Be 2+ is isoelectronic with which of the following ions ?
A H +
B Li +
C Na +
D Mg 2+
Correct Answer
Option B
Solution

Be 2+ has 2 electrons and also Li + has 2 electrons so Be 2+ and Li + are isoelectronic

Q25
What is the maximum number of orbitals that can be identified with the following quantum numbers ? n = 3, l = 1, m 1 = 0
A 1
B 2
C 3
D 4
Correct Answer
Option A
Solution

Only one orbital 3p has following set of quantum numbers, n = 3, l = 1 and m l = 0

Q26
According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 × \times 10 −-27 ergs, c = 3 × \times 10 10 cm s −-1 , N A = 6.02 × \times 10 −-23 mol −-1 )
A 1.196×108λ{{1.196 \times {{10}^8}} \over \lambda }
B 2.859×105λ{{2.859 \times {{10}^5}} \over \lambda }
C 2.859×1016λ{{2.859 \times {{10}^{16}}} \over \lambda }
D 1.196×1016λ{{1.196 \times {{10}^{16}}} \over \lambda }
Correct Answer
Option A
Solution

E =

hcNAλ{{hc{N_A}} \over \lambda }

=

6.62×10−27×3×1010×6.02×1023λ{{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}} \over \lambda }

=

1.1955×108λ{{1.1955 \times {{10}^8}} \over \lambda }

=

1.196×108λ{{1.196 \times {{10}^8}} \over \lambda }

ergs mol -1

Q27
The outer electronic configuration of Gd (At. No. 64) is
A 4f 5 5d 4 6s 1
B 4f 7 5d 1 6s 2
C 4f 3 5d 5 6s 2
D 4f 4 5d 5 6s 1
Correct Answer
Option B
Solution

Gd[64] = [Xe]4f 7 5d 1 6s 2

Q28
What is the maximum numbers of electrons that can be associated with the following set of quantum numbers ? n = 3, ll = 1 and m = −- 1
A 4
B 2
C 10
D 6
Correct Answer
Option B
Solution

The orbitals associated with n = 3, l = 1 is 3p.

One orbital (with m = -1) of 3p-subshell can only accomodate maximum 2 electrons.

Q29
Based on equation E = −- 2.178 × \times 10 −-18 J (Z2n2)\left( {{{{Z^2}} \over {{n^2}}}} \right), certain conclusions are written. Which of them is not correct ?
A Equation can be used to calculate the change in energy when the electron changes orbit.
B For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.
C The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
D Larger the value of n, the larger is the orbit radius.
Correct Answer
Option B
Solution

The electron is more tightly bound in the smallest allowed orbit.

Q30
The value of Planck's constant is 6.63 × \times 10 −-34 J s. The speed of light is 3 × \times 10 17 mm s −-1 . Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 × \times 10 15 s −-1 ?
A 50
B 75
C 10
D 25
Correct Answer
Option A
Solution

c =

νλ\nu \lambda

λ\lambda =

cν{c \over \nu }

=

3×10176×1015{{3 \times {{10}^{17}}} \over {6 \times {{10}^{15}}}}

= 50 nm

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