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Periodic Properties (NEST)

Description: Inorganic chemistry Classification of Elements and Periodicity in Properties
Number of Questions: 15
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Tags: Inorganic chemistry Classification of Elements and Periodicity in Properties
Attempted 0/15 Correct 0 Score 0

Arrange the following in increasing order of H-E-H bond angle:

  1. NH3 2. PH3 3. AsH3 4. SbH3
  1. 1 < 2 < 3 < 4

  2. 2 < 4 < 3 < 1

  3. 3 < 2 < 1 < 4

  4. 4 < 2 < 3 < 1

  5. 4 < 3 < 2 < 1


Correct Option: E
Explanation:

All of these molecules have pyramidal geometry. On moving down the group, H-E-H bond angle decreases due to increase in atomic size. So, correct order of increasing H-E-H bond angles is SbH3 < AsH3 < PH3 < NH3.

Which of the following trends of properties is incorrect?

  1. Reducing nature: HF < HCl < HBr < HI

  2. Bond length: HF < HCl < HBr < HI

  3. Acidic strength: HF < HCl < HBr < HI

  4. Dipole moment: HF < HCl < HBr < HI

  5. All of the above are correct.


Correct Option: D
Explanation:

Dipole moment of hydrogen halides decreases down the group with decrease in electronegativty of halogens. Thus, the correct order of dipole moment is HF > HCl > HBr > HI.

Which among the following properties follows the trend H2O > H2S > H2Se > H2Te?

  1. Acidic strength and reducing nature

  2. Bond angle and thermal stability

  3. Acidic strength and thermal stability

  4. Acidic strength and bond angle

  5. Bond angle and reducing nature


Correct Option: B
Explanation:

H-E-H bond angle as well as thermal stability of group 16 hydrides decrease down the group, i.e. H2O > H2S > H2Se > H2Te.

The correct order of ease of liquefaction of noble gases is

  1. Xe > Ne > Ar > Kr > He

  2. He > Ne > Ar > Kr > Xe

  3. He < Ar < Kr < Ne < Xe

  4. He < Ne < Ar < Kr < Xe

  5. Ne < Ar < He < Kr < Xe


Correct Option: D
Explanation:

Ease of liquefaction of noble gases increases down the group. So, correct order is He < Ne < Ar < Kr < Xe.

Identify the TRUE statement(s).

  1. Ni(II) compounds are thermodynamically less stable than Pt(II) compounds.
  2. Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds.
  3. Mn(II) compounds are less stable than Fe(II) towards oxidation to their +3 state.
  1. Only 1

  2. Only 2

  3. Only 3

  4. Both 1 and 2

  5. Both 2 and 3


Correct Option: B
Explanation:

The sum of first four ionisation enthalpies (IE1 + IE2 + IE3 + IE4) for Pt is lesser than that of nickel. Therefore, it requires a smaller amount of energy to ionise Pt to Pt4+ ion. Thus, Pt(IV) compounds are thermodynamically more stable than Ni(IV) compounds.

Arrange the following s-block elements in increasing order of melting point:

  1. Sodium
  2. Potassium
  3. Magnesium
  4. Calcium
  1. 1 < 2 < 3 < 4

  2. 2 < 1 < 3 < 4

  3. 2 < 3 < 1 < 4

  4. 3 < 2 < 4 < 1

  5. 4 < 2 < 1 < 3


Correct Option: B
Explanation:

Correct order of melting points is potassium (336 K) < sodium (371 K) < magnesium (924 K) < calcium (1124 K).

Which of the following statements are FALSE?

  1. Thermal stability of alkaline earth metal carbonate increases down the group.
  2. Thermal stability of alkaline earth metal oxides decreases down the group.
  3. Solubility of alkaline earth metal hydroxides in water decreases down the group.
  4. Solubility of alkaline earth metal carbonates in water increases down the group.
  1. 1 and 2

  2. 1 and 4

  3. 2 and 3

  4. 2 and 4

  5. 3 and 4


Correct Option: E
Explanation:

Both are false statements. Among alkaline earth metal hydroxides, the anion being common the cationic radius will influence the lattice enthalpy. Since lattice enthalpy decreases much more than the hydration enthalpy with increasing ionic size, the solubility increases on moving down the group. The size of anions being much larger compared to cations, the lattice enthalpy will remain almost constant within a particular group. Since the hydration enthalpies decrease down the group, solubility of alkaline earth metal carbonates will decrease.

The increasing order of effective nuclear charge is

  1. Na < Mg < Al < Si

  2. Mg < Na < Al < Si

  3. Mg < Mg < Si < Al

  4. Al < Mg < Na < Si

  5. Si < Al < Mg < Na


Correct Option: A
Explanation:

On moving from left to right in a period, effective nuclear charge increases with increase in atomic number. Thus, correct order of effective nuclear charge is Na < Mg < Al < Si.

Correct order of increasing dipole moment of HF, HCl, HBr and HI is

  1. HF< HCl < HBr < HI

  2. HI< HCl < HBr < HF

  3. HI< HCl < HF < HBr

  4. HI< HBr < HCl < HF

  5. HCl< HF < HBr < HI


Correct Option: D
Explanation:

With increase in electronegativity, difference between the bonded atoms, dipole moment increases. On moving down the group, electronegativity decreases. So, the correct order is HI< HBr < HCl < HF.

Which among the following transition metal ions are colourless?

  1. Sc3+
  2. Mn2+
    3. Hg2+ 4. Fe3+ 5. Cu+
  1. 1, 2 and 3

  2. 1, 3 and 5

  3. 2, 3 and 4

  4. 2, 4 and 5

  5. 3, 4 and 5


Correct Option: B
Explanation:

Transition metal ions exhibit colour due to the presence of unpaired electrons in (n−1) d-orbitals which undergo d-d transition. The metal ions not having unpaired electrons (n−1)d0 or having completely filled d orbital (n−1)d10 do not absorb radiations in visible region, since d-d transitions are not possible. Hence, they are colourless ions. Sc3+ (3d0), Cu+ (3d10), and Hg2+ (5d10): All are colourless

Correct order of acidic character of oxides of manganese is

  1. MnO < MnO2 < Mn2O3 < Mn2O7

  2. Mn2O7 < Mn2O3 < MnO2 < MnO

  3. MnO < Mn2O3 < MnO2 < Mn2O7

  4. MnO < Mn2O3 < Mn2O7 < MnO2

  5. Mn2O3 < MnO2 < MnO < Mn2O7


Correct Option: D
Explanation:

Correct order of acidic behaviour is Mangnese(II) oxide < Manganese(III) oxide < Manganese(IV) oxide < Manganese(VII) oxide.

Which of the following matches of 'element: electronic configuration' is incorrect?

  1. Cerium: [Xe] 4f2 5d0 6s2

  2. Samarium: [Xe] 4f6 5d0 6s2

  3. Gadolinium: [Xe] 4f8 5d0 6s2

  4. Holmium: [Xe] 4f11 5d0 6s2

  5. Lutetium: [Xe] 4f14 5d1 6s2


Correct Option: C
Explanation:

Correct electronic configuration of gadolinium (64Gd) is [Xe] 4f7 5d1 6s2.

Correct order of increasing values of second ionisation potential of C, N, O and F is

  1. C < N < O < F

  2. C < O < N < F

  3. C < N < F < O

  4. O < N < F < C

  5. F < O < N < C


Correct Option: C
Explanation:

The second ionisation potential means energy required for removal of electron from following cations: C+ (5e) = 1s2, 2s2, 2p1 N+ (6e) = 1s2, 2s2, 2p2 O+ (7e) = 1s2, 2s2, 2p3 F+ (8e) = 1s2, 2s2, 2p4 Therefore, order of increasing second ionisation potential is C < N < F < O.

Correct order of decreasing ionic size for K+, Ca2+, Cl and S-- is

  1. K+ > Ca++ > Cl- > S--

  2. Ca++ > K+ > Cl > S--

  3. K+ > Ca++ > Cl > S--

  4. Cl > S-- > Ca++ > K+

  5. S-- > Cl > K+ > Ca++


Correct Option: E
Explanation:

All are isoelectronic species having twenty electrons each. In isoelectronic species, ionic size decreases with increase in atomic number. So, the correct order of ionic size is S-- > Cl– > K+ > Ca++.

Assertion (A): HClO4 is a stronger acid than HClO3. Reason (R): ClO4- ion is more stable than ClO3-.

Directions: In the following question there are two statements, Assertion (A) and Reason (R). Consider both the statements independently and mark your answers as under:

1. If A and R both are correct and R is the correct explanation of A.
2. If A and R both are correct and R is not the correct explanation of A.
3. If A is correct and R is incorrect.
4. If A is incorrect and R is correct.
5. If A and R both are incorrect.

  1. 1

  2. 2

  3. 3

  4. 4

  5. 5


Correct Option: A
Explanation:

HClO4 is a stronger acid than HClO3. This can be explained on the basis of relative stability of the anion or conjugate base formed after removal of a proton. ClO4- is more stable than ClO3- due to greater dispersal of negative charge on four oxygen atoms. So, HClO4 is a stronger acid than HClO3.

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