The increasing order of the ionic radii of the given isoelectronic species is :

Cl^-, Ca²⁺, K⁺, S^2-
S^2-, Cl^-, Ca²⁺, K⁺
Ca²⁺, K⁺, Cl^-, S^2-
K⁺, S^2-, Ca²⁺, Cl^-

Solution

Among isoelectronic species ionic radii increases as the charge increases.

Order of ionic radii Ca²⁺ < K⁺ < Cl^- < S^2-

The number of electrons remains the same but nuclear charge increases within crease in the atomic number causing decrease in size.

The frequency of light emitted for the transition n = 4 to n = 2 of the He⁺ is equal to the transition in H atom corresponding to which of the following ?

n = 2 to n = 1
n = 3 to n = 2
n = 4 to n = 3
n = 3 to n = 1

Solution

DIRECTIONS : These are Assertion-Reason type questions. Each of these question contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Answer these questions from the following four options.

Statement – 1 : Nuclide ³⁰Al₁₃ is less stable than ⁴⁰Ca₂₀

Statement – 2 : Nuclide having odd number of protons and neutrons are generally unstable.

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is False
Statement - 1 is False, Statement - 2 is True

Solution

It is observed that a nucleus which is made up of even number of nucleons (No. of n & p) is more stable than nuclie which consist of odd number of nucleons. If number of neutron or proton is equal to some numbers i.e., 2, 8, 20, 50, 82, or 126 (which are called magic numbers),then these posses extra stability.

Statement – 1: The radius of the first orbit of hydrogen atom is 0.529Å.

Statement – 2 : Radius of each circular orbit (r_n) – 0.529Å (n²/Z), where n = 1, 2, 3 and Z = atomic number.

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is False
Statement - 1 is False, Statement - 2 is True

Solution

Statement – 1 : The position of an electron can be determined exactly with the help of an electron microscope.

Statement – 2 : The product of uncertainty in the measurement of its momentum and the uncertainty in the measurement of the position cannot be less than a finite limit.

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
Statement - 1 is True, Statement - 2 is False
Statement - 1 is False, Statement - 2 is True

Solution

The statement-1 is false but the statement-2 is true exact position and exact momentum of an electron can never be determined according to Heisenberg’s uncertainty principle. Even not with the help of electron microscope because when electron beam of electron microscope strikes the target electron of atom, the impact causes the change in velocity and position of electron. Thus the product of uncertainty in position and momentum is

Δx.Δp ≥ ^h⁄_4π ≈ 0.57 ergs sec/gram

Energy of an electron is given by E = –2.178 × 10^-18 J(^Z²⁄_n²)Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be :(h = 6.62 × 10^-34Js and c = 3.0 × 10⁸ms^-1)

1.214 × 10^-7 m
2.816 × 10^-7 m
6.500 × 10^-7 m
8.500 × 10^-7 m

Solution

Given that the abundances of isotopes ⁵⁴Fe, ⁵⁶Fe and ⁵⁷Fe are 5%, 90% and 5%, respectively, the atomic mass of Fe is

55.85
55.95
55.75
56.05

Solution

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a₀ is Bohr radius] :

\(\frac{h^{2}}{4 \pi^{2}ma_{0}^{2}}\)
\(\frac{h^{2}}{16 \pi^{2}ma_{0}^{2}}\)
\(\frac{h^{2}}{32 \pi^{2}ma_{0}^{2}}\)
\(\frac{h^{2}}{64 \pi^{2}ma_{0}^{2}}\)

Solution

The electrons identified by quantum numbers n and l :

(A) n = 4, l = 1

(B) n = 4, l = 0

(D) n = 3, l = 1

can be placed in order of increasing energy as :

(C) < (D) < (B) < (A)
(D) < (B) < (C) < (A)
(B) < (D) < (A) < (C)
(A) < (C) < (B) < (D)

Solution

(A) 4 p
(B) 4 s
(C) 3 d
(D) 3 p

According to Bohr Bury's (n + l) rule, increasing order of energy will be (D) < (B) < (C) < (A). Note : If the two orbitals have same value of (n + l)then the orbital with lower value of n will be filled first.

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