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Trends in the periodic table - class-XI

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On the Mulliken scale, the average of ionization potential and electrons affinity is known as _____(electronegativity/affinity).

  1. electronegativity

  2. electroneaffinity

  3. electropositivity

  4. none


Correct Option: A
Explanation:

On Mulliken scale, the average of ionization potential and electrons affinity is known as electronegativity.

If the E.N. value of Nitrogen in the pauling scale is 3.0, then he value in mulliken scale will be

  1. $11.2$

  2. $22.4$

  3. $8.4$

  4. $3.0$


Correct Option: C

Which of the following is the correct order of hydrated radii?

  1. $Li^{+} < Na^{+} < K^{+} < Rb^{+} < Cs^{+}$

  2. $Rb^{+} < Na^{+} < Li ^{+} < Cs^{+} < K^{+}$

  3. $Cs^{+} < Rb^{+} < K^{+} < Na ^{+}< Li^{+}$

  4. $Li^{+} < K^{+} < Na^{+} < Rb ^{+} < Cs ^{+}$


Correct Option: A

Electronegativity of $F$ on Pauling's scale is 4.0. The value on Mulliken's scale is:

  1. $11.2$

  2. $14.4$

  3. $16.8$

  4. None of these


Correct Option: A
Explanation:

Mulliken proposed that the arithmetic mean of the first ionization energy (I) and the electron affinity (Eea) should be a measure of the tendency of an atom to attract electrons.
$X= \dfrac{E _i + E _{Ea}}{2}$
Electronegativity of $F$ on Mulliken's scale is hence $11.2$

EN(mulliken scale) of the element (A) is $E _1$ and IP is $E _2$. Hence, EA will be:

  1. $2E _1 - E _2$

  2. $E _1 - E _2$

  3. $E _1 - 2E _2$

  4. $(E _1 + E _2)/2$


Correct Option: A
Explanation:

As we know,
$\displaystyle EN = \frac{(IP) + (EA)}{2}$ 


$EA = 2 EN - IP$

$EA = 2E _1 - E _2$

The ionic radii of ${Rb}^{+}$ and ${I}^{-}$ are $1.46$ and $2.16$ $\mathring { A } $, respectively. The most probable type of structure exhibited by it is

  1. $CsCl$ type

  2. $NaCl$ type

  3. $ZnS$ type

  4. $Ca{F} _{2}$ type


Correct Option: B

Which pair of element has almost identical electronegativity values?

  1. $In,Tl$

  2. $B,Al$

  3. $B,Ga$

  4. $Al,Tl$


Correct Option: A
Explanation:

Both have close IE and electron affinity values(both are metals) and hence similar electronegativity values.
$Electronegativity=\dfrac{IE+EA}{2}$

Mulliken scale of elecrtonegativity uses the concept of:

  1. E. A. and EN of pauling

  2. E.A. and atomic size

  3. E.A. and I.P.

  4. E.A. and bond energy


Correct Option: C
Explanation:

Mulliken proposed that the arithmetic mean of the first ionization energy and the electron affinity should be a measure of the tendency of an atom to attract electrons.

If the ionisation potential is IP, electron affinity is EA and electronegativity is x then which of the following relation is correct.

  1. 2X-EA-IP=0

  2. 2EA-X-IP=0

  3. 2IP-X-EA=0

  4. All of the above


Correct Option: A
Explanation:
Electron affinity: The amount of energy released when an electron is added to the outermost shell of one mole of an isolated gaseous atom in its lower energy state.

Ionisation energy: Minimum energy required to remove most loosly held outer most shell e – in ground state from an isolated gaseous atom is known as ionisation potential.

$\displaystyle X= \frac{IP+EA}{2}$

$2X-IP-EA=0$

Electronegativity of F on Pauling scale is 4.0. Its value on Mulliken scale is__________.

  1. 12.35

  2. 11.0

  3. 15.2

  4. 14.2


Correct Option: A
Explanation:

Electronegativity of F on Pauling scale is 4.0. Its value on Mulliken scale is 11.2
$\displaystyle \chi (Pauling) = 0.34 \chi (Mulliken) -0.2 $
$\displaystyle 4.0 = 0.34\chi (Mulliken) -0.2 $
$\displaystyle \chi (Mulliken) = 12.35 $

Define the relation among the $EN,\,IE\;and\;E.A.$

  1. $EN=\displaystyle\frac{IE\times EA}{2}$

  2. $EN=IE+EA$

  3. $2EN=IE-EA$

  4. $EN=\displaystyle\frac{IE+EA}{2}$

  5. $EN=\displaystyle\frac{IE-EA}{2}$


Correct Option: D
Explanation:

The correct relationship is $EN = \dfrac {IE+EA}{2}$
According to mulliken's scale, the electronegatiity is the average value of ionization potential and electron affinity of an atom.

The $Z _{effective}$ for $He$ is?

  1. 2

  2. 1.7

  3. 1.85

  4. 1.65


Correct Option: A
Explanation:
The effective nuclear charge experienced by a 1s electron in helium is +1.70.

The effective nuclear charge $Z _{eff}$ is the net positive charge experienced by an electron in a multi-electron atom.

A given electron does not experience a full nuclear charge because the other electrons are sometimes between it and the nucleus and shield it from the nucleus.

The formula for effective nuclear charge is-

$Z _{eff}=Z-S$

where, 
is the number of protons in the nucleus, and S is the shielding constant, the average number of electrons between the nucleus and the electron in question.
The American physicist John Slater derived a number of rules to determine the shielding constant.

He found that for electrons in a 1s orbital, the second electron shields the first by 0.30 units.

$Z _{eff}=Z-S=2- 0.30-1.70$

Hence, the correct option is B.


The screening effect of 'd' electrons is : 

  1. much more than s-electrons

  2. equal to s-electrons

  3. equal to p-electrons

  4. much less than s-electrons


Correct Option: D

An atom of each element has a definite combining capacity called : 

  1. valency

  2. affinity

  3. bonding

  4. energy levels


Correct Option: A
Explanation:

An atom of each element has a definite combining capacity called its valency.

For example Na has electronic configuration 2,8,1 
So, valency is 1 as after losing one electron it will have stable octet.

The electrons present in the outermost shell of an atom are known as :

  1. protons

  2. neutrons

  3. octet

  4. valence electrons


Correct Option: D
Explanation:

The electrons present in the outermost shell of an atom are known as valence electrons.

The valency of nitrogen in nitrogen dioxide is:

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: D
Explanation:

$(D)$  $ 4$


$Sol. $  let the valency of nitrogen in be $ x$. 
           valency of oxygen = $ -2 $
          For $NO _2 $ it will be ,
                                $ x + 2 \times(-2) = 0 $
                                 $ x = +4 $
Hence, the valency of nitrogen in $ NO _2$ is $ +4 $. 

Valency of magnesium and oxygen in $MgO$ is:

  1. one and one

  2. two and two

  3. one and two

  4. two and one


Correct Option: B
Explanation:

Ans. $(B)$ $(two$ $and$ $two)$

This is because magnesium loses two electrons to have an octet and oxygen gains two electrons to have an octet. The final formula of magnesium oxide is ${MgO}$. So, the valencies magnesium and oxygen in $ {MgO}$ are $2$ and $2$ .

Valence electrons are responsible for the chemical properties of an atom.

  1. True

  2. False


Correct Option: A

The element with the highest first ionisation potential is:

  1. Boron

  2. Carbon

  3. Nitrogen

  4. Oxygen


Correct Option: C

The ionisation potential of hydrogen atom is $13.6\ eV$. The energy of required to remove an electrons in the $n=2$ state of hydrogen atom is:

  1. $27.2\ eV$

  2. $13.6\ eV$

  3. $6.8\ eV$

  4. $3.4\ eV$


Correct Option: D
Explanation:
$E _n=-13.6\ ev\left (\dfrac {Z^2}{n^2}\right)=\dfrac {-E _1}{n^2}[I.E _1 =-E _1 =13.6\ eV \Rightarrow E _1=-13.6\ eV]$

$\therefore E _2 =\dfrac {-13.6\ ev}{4}=-3.4ev$

$\therefore I.E _2=-E _2 =3.4\ ev$.

Option D is correct.
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