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Periodic Table & Periodicity

NEET Chemistry · 94 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
Which one of the following orders is not in accordance with the property stated against it?
A F 2 > Cl 2 > Br 2 > I 2 : Bond dissociation energy
B F 2 > Cl 2 > Br 2 > I 2 : Oxidising power
C HI > HBr > HCl > HF : Acidic property in water
D F 2 > Cl 2 > Br 2 > I 2 : Electronegativity.
Correct Answer
Option A
Solution

Bond dissociation energy is kcal/mol of F - F, Cl - Cl, Br - Br and I - I are 38, 57, 45.5 and 35.6 respectively.

Bond dissociation energy of fluorine is less because of its small size and repulsion between electrons of two atoms.

Q32
Which one of the following orders is not in accordance with the property stated against it ?
A F 2 > Cl 2 > Br 2 > I 2 : Bond dissociation energy
B F 2 > Cl 2 > Br 2 > I 2 : Oxidising power
C HI > HBr > HCl > HF : Acidic property in water
D F 2 > Cl 2 > Br 2 > I 2 : Electronegativity
Correct Answer
Option A
Solution
F−F38   Cl−Cl57  Br−Br45.5  I−I35.6\mathop {F - F}\limits_{38} \,\,\,\mathop {Cl - Cl}\limits_{57} \,\,\mathop {Br - Br}\limits_{45.5} \,\,\mathop {I - I}\limits_{35.6}

X - X bond dissociation energy (kcal/mol) The lower value of bond dissociation energy of fluorine is due to the high inter-electronic repulsion between non-bonding electrons in the 2p-orbitals of fluorine.

As a result F - F bond is weaker in comparison to Cl - Cl and Br - Br bonds.

Q33
Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species ?
A S < O < Cl < F
B Cl < F < S < O
C F < Cl < O < S
D O < S < F < Cl
Correct Answer
Option D
Solution

The molar enthalpy change accompanying the addition of an electron to an atom (or ion) is known as electron gain enthalpy.

Generally in increases on moving from left to right in a period and in a group it decreases as the size increases.

Exception: Because of the small size of F, electron-electron repulsion present in its relatively compact 2p-subshell, do not easily allow the addition of an extra electron.

On the other hand, Cl because of its comparatively bigger size than F, allows the addition of an extra electron more easily.

O−1.48<S−2.0<F−3.6<Cl−3.8\mathop O\limits_{ - 1.48} < \mathop S\limits_{ - 2.0} < \mathop F\limits_{ - 3.6} < \mathop {Cl}\limits_{ - 3.8}
Q34
Ionic radii are
A inversely proportional to effective nuclear charge
B inversely proportional to square of effective nuclear charge
C directly proportional to effective nuclear charge
D directly proportional to square of effective nuclear charge.
Correct Answer
Option A
Solution

Ionic radius in the n th orbit is given by r n =

n2a0Z∗{{{n^2}{a_0}} \over {{Z^*}}}

or,

rn∝1Z∗{r_n} \propto {1 \over {{Z^*}}}

When n is the principal quantum number, a 0 the Bohr's radius of H-atom and Z * , the effective nuclear charge.

Q35
The ions O 2−- , F −- , Na + , Mg 2+ and Al 3+ are isoelectronic. Their ionic radii show
A a significant increase from O 2−- to Al 3+
B a significant decrease from O 2−- to Al 3+
C an increase from O2 −- to F −- and then decrease from Na + to Al 3+
D a decrease from O2 −- to F −- and then increase from Na + to Al 3+ .
Correct Answer
Option B
Solution

Amongst isoelectronic ions, ionic radii of anions is more than that of cations.

Further size of the anion increases with increase in -ve charge and size of cation decreases with increase in +ve charge.

Hence, the correct order is O 2- > F - > Na + > Mg 2+ > Al 3+

Q36
Which of the following order is wrong?
A NH 3 < PH 3 < AsH 3 −- acidic
B Li < Be < B < C −- 1 st IP
C Al 2 O 3 < MgO < Na 2 O < K 2 O −- basic
D Li + < Na + < K + < Cs + −- ionic radius.
Correct Answer
Option B
Solution

Along the period, I.P. generally increases but not regularly.

Be and B are exceptions.

First I.P. increases in moving from left to right in a period, but I.P. of B is lower than Be.

As in case of boron, 2p 1 electron have to be removed to get B + (1s 2 2s 2 ) from B(1s 2 2s 2 2p 1 ), while in case of Be (1s 2 2s 2 ) electron have to be removed to get Be + (1s 2 2s 1 ). p electron can be removed more easily than s electron so the energy required to remove electron will be less in case of boron.

Q37
Correct order of 1 st ionization potential among following elements Be, B, C, N, O is
A B < Be < C < O < N
B B < Be < C < N < O
C Be < B < C < N < O
D Be < B < C < O < N
Correct Answer
Option A
Solution

Be – 1s 2 2s 2 ; B – 1s 2 2s 2 2p 1 ; C – 1s 2 2s 2 2p 2 ; N – 1s 2 2s 2 2p 3 ; O – 1s 2 2s 2 2p 4 .

IP increases along the period.

But IP of Be > B.

Further IP of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy.

As in case of boron, 2p 1 electron have to be removed to get B + (1s 2 2s 2 ) from B(1s 2 2s 2 2p 1 ), while in case of Be (1s 2 2s 2 ) electron have to be removed to get Be + (1s 2 2s 1 ). p electron can be removed more easily than s electron so the energy required to remove electron will be less in case of boron.

Q38
Which statement is wrong ?
A Bond energy of F 2 > Cl 2
B Electronegativity of F > Cl
C F is more oxidising than Cl
D Electron affinity of Cl > F
Correct Answer
Option A
Solution

Bond energy generally decreases on moving from top to bottom along a group.

It happens due to fact that size increases on moving down the group and thus, the two nuclei are far apart and less capable of holding the two atoms together.

But in case of F 2 molecule due to more repulsion in between non-bonding electron pair (2p) of two fluorines (due to small size of F-atom) in comparison to non-bonding electron pair (3p) in chlorine, the bond energy of F 2 is less than Cl 2 .

BE(F 2 ) = 158.5 kJ/mole and BE(Cl 2 )= 242.6 kJ/mole

Q39
In which of the following options the order of arrangement does not agree with the variation of property indicated against it ?
A I < Br < Cl < F (increasing electron gain enthalpy)
B Li < Na < K < Rb (increasing metallic radius)
C Al 3+ < Mg 2+ < Na + < F −- (increasing ionic size)
D B < C < N < O (increasing first ionisation enthalpy)
Correct Answer
Option A
Solution

The correct order of increasing negative electron gain enthalpy is : I < Br < F < Cl and the correct order of increasing first ionisation enthalpy is B < C < O < N.

Q40
Which one of the following arrangements does not give the correct picture of the trends indicated against it ?
A F 2 > Cl 2 > Br 2 > I 2 : Bond dissociation energy
B F 2 > Cl 2 > Br 2 > I 2 : Electronegativity
C F 2 > Cl 2 > Br 2 > I 2 : Oxidizing power
D F 2 > Cl 2 > Br 2 > I 2 : Electron gain enthalpy
Correct Answer
Option A
Solution

For Diatomic molecules (X 2 ) of halogens the bond dissociation energy decreases in the order: Cl 2 > Br 2 > F 2 > I 2 The oxidising power, electronegativity and relativity decrease in the order: F 2 > Cl 2 > Br 2 > I 2 Order of Electron gain enthalpy of halogens Cl 2 > F 2 > Br 2 > I 2 The low value of electron gain enthalpy of fluorine is probably due to small size of fluorine atom.

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