The wavelength of the radiation emitted, when in a hydrogen atom

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

Which one of the following compounds has the smallest bond angle in its molecule ?

  • SO2

  • OH2

  • SH2

  • NH3


2.

The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10-5 mol L-1. Its solubility product number will be

  • 4 × 10-15

  • 4 × 10-10

  • 1 × 10-15

  • 1 × 10-10


3.

If at 298 K the bond energies of C-H, C-C, C=C and H-H bonds are respectively 414, 347, 615 and 435 kJ mol-1, the value of enthalpy change for the reaction

H2C=CH2 (g) + H2 (g) → H3C - CH3 (g) at 298 K will be

  • +250 kJ

  • -250 kJ

  • +125 kJ

  • -125 kJ


4.

The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2NH is

  • CH3NH2 < NH3 < (CH3)2NH

  • (CH3)2NH < NH3 < CH3NH2

  • NH3 < CH3NH< (CH3)2NH

  • CH3NH< (CH3)2NH < NH3


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5.

The solubilities of carbonates decrease down the magnesium group due to a decrease in

  • lattice energies of solids

  • hydration energies of cations

  • inter-ionic attraction

  • entropy of solution formation


6.

Due to the presence of an unpaired electron free radicals are

  • cations

  • anions

  • chemically inactive

  • chemically reactive


7.

Consider the reaction,

N2 + 3H2 → 2NH3

carried out at constant temperature and pressure. If H and U are the enthalpy and internal energy changes for the reaction, which of the following expression is true?

  • H > U

  • H <U

  • H = U

  • H = 0


8.

The molecular shapes of SF4, CF4 and XeF4 are

  • different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively

  • different with 0, 1 and 2 lone pairs of electrons on the central atom, respectively

  • the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively

  • the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively


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9.

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

  • 91 nm

  • 192 nm

  • 406 nm

  • 9.1 × 10-8 nm


A.

91 nm

   1λ = vH = RH 1n12 - 1n22       = 1.097 × 107 112 - 1 λ = 11.097 × 107 m

      = 9.11 × 10-8 m

      = 91.1 × 10-9 m

      = 91.1 nm  


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10.

What is the equilibrium expression for the reaction

P4 (s) + 5O2 (g)  P4O10 (s) ?

  • KcP4O10[P4][O2]5

  • KcP4O105[P4][O2]

  • Kc = [O2]5

  • Kc1[O2]5


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