The number of lone pairs of electrons present on : Xe in XeF4 is

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The electronic configuration of elements A, B and C are [He] 2s1, [Ne] 3sand [Ar] 4s1 respectively, which one of the following order is correct for the first ionization potentials (in kJ mol-1) of A , B and C?

  • A > B > C

  • C > B > A

  • B > C > A

  • C > A > B


2.

The catenation tendency of C, Si and Ge is in the order Ge < Si < C. The bond energies (in kJ mol-1) of C-C, Si-Si and Ge-Ge bonds, respectively are

  • 167, 180, 348

  • 180, 167, 348

  • 348, 167, 180

  • 348, 180, 167


3.

Ionic radius (in Å) of As3+, Sb3+ and Bi3+ follow the order

  • As3+ > Sb3+ > Bi3+

  • Sb3+ > Bi3+ > As3+ 

  • Bi3+ > As3+ > Sb3+

  • Bi3+ > Sb3+ > As3+


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4.

The number of lone pairs of electrons present on : Xe in XeF4 is

  • 3

  • 4

  • 1

  • 2


D.

2

Xe in ground state → 5s2, 5p6, 5d0

Xe in IInd excited state → 5s2, 5p4, 5d2

Number of lone pairs = 2

Number of bond pairs = 4

Therefore, the number of lone pairs of electrons present on Xe in XeF4 are 2.


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5.

Which one of the following ions exhibits highest magnetic moment ?

  • Cu2+

  • Ti3+

  • Ni2+

  • Mn2+


6.

The energy of an electron present in Bohr's second orbit of hydrogen atom is

  • -1312 J atom-1

  • -328 kJ mol-1

  • -328 J mol-1

  • -164 kJ mol-1


7.

In the ground state, an element has 13 electrons in M shell. The element is

  • copper

  • chromium

  • nickel

  • iron


8.

In a nuclide, one a.m.u. of mass is dissipated into energy to bind its nucleons. The energy equivalent of this mass is

  • 931.5 eV

  • 931.5 × 106 eV

  • 931.5 × 106 MeV

  • 931.5 MV


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9.

What is X in the following reaction ?

12Mg261H2 → 12Mg27 + X

  • γ-ray

  • 0n1

  • 1H1

  • 1D2


10.

Let electronegativity, ionisation energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?

  • EN = IP × EA

  • EN = IPEA

  • EN= IP+ EA2

  • EN = IP - EA


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