The “spin-only” magnetic moment [in units of Bohr magneton,(µB )] of Ni²⁺ in aqueous solution would be (Atomic number of Ni = 28) 

• 2.84

• 4.90

• 1

• 1

Which of the following sets of quantum numbers is correct for an electron in 4f orbital?

• n = 4, I =3, m = +4, s = + 1/ 2

• n = 3, I = 2, m = -2, S = + 1/2

• n =4, I = 3, m = +1, s = + 1/ 2

• n =4, I = 3, m = +1, s = + 1/ 2

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers I =1 and 2 are respectively

• 12 and 4

• 16 and 5

• 16 and 4

• 16 and 4

The wavelength of the radiation emitted, when in hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×10⁷ m^-1)

• 91 nm

• 9.1×10^-8 nm

• 406 nm

• 406 nm

If the given four electronic configurations,

(i) n = 4, l = 1

(ii) n = 4, l = 0

(iii) n = 3, l = 2

(iv) n = 3, l = 1

are arranged in order of increasing energy, then the order will be

• (iv) < (ii) < (iii) < (i)

• (ii) < (iv) < (i) < (iii)

• (i) < (iii) < (ii) < (iv)

• (iii) < (i) < (iv) < (ii)

Which of the following sets of quantum numbers represents the 19^th electron of Cr(Z = 24)?

• (4, 1, -1, +$\frac{1}{2}$)

• (4, 0, 0, +$\frac{1}{2}$)

• (3, 2, 0, -$\frac{1}{2}$)

• (3, 2, -2, +$\frac{1}{2}$)

C₆H₅F¹⁸ is a F¹⁸  radio-isotope labelled organic compound. F¹⁸ decays by positron emission. The product resulting on decay is 

• C₆H₅O¹⁸

• C₆H₅Ar¹⁹

• B¹²C₅H₅F

• C₆H₅O¹⁶

278.

The energy required to break one mole of hydrogen-hydrogen bonds in H₂ is 436 kJ. What is the longest wavelength of light required to break a single hydrogen-hydrogen bond?

• 68.5nm

• 137nm

• 274nm

• 584nm

An element E loses one $\mathrm{\alpha }$ and two $\mathrm{\beta }$ particles in three successive stages. The resulting elemenl wiil be

• an isobar of E

• an isotone of E

• an isotope of E

• E itself

Which one of the following corresponds to a photon of highest energy?

• λ = 300 mm

• v = 3 × 10⁸ s^-1

• $\overline{\mathrm{v}}$ = 30 cm^-1

• ε = 6.626 × 10^-27 J