Tag: electron configuration

Questions Related to electron configuration

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

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. $EN=\displaystyle\frac{IE-EA}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

An element $X$ has $IP = 1681$ kJ/mole and $EA =-333$ kJ/mole then its electronegativity is:

  1. $(1681 + 333) / 544$
  2. $(1681 - 333 )/ 544$
  3. $(1681 + (-333)) / 2$
  4. $\dfrac{208\sqrt{1681+333}}{544}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Robert S. Mulliken proposed that the arithmetic mean of the first ionization energy ($E i$) and the electron affinity ($E _{ea}$) should be a measure of the tendency of an atom to attract electrons. As this definition is not dependent on an arbitrary relative scale, it has also been termed absolute electronegativity, with the units of kilojoules per mole or electron volts.
                                        <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/44e7db9c2cb1be249c807e9435c4d11788ea2efc" class="mwe-math-fallback-image-inline" alt="\chi =(E
{\rm {i}}+E_{\rm {ea}})/2\,">

Multiple choice chemistry classification of elements- the periodic table periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

The ${ Z } { eff }$ for
3d electron of Cr
4s electron of Cr
3d electron of ${ Cr }^{ 3+ }$
3s electron of ${ Cr }^{ 3+ }$ are _
________ respectively.

  1. 4.6, 2.95, 4.95, 8.05

  2. 4.95, 2.05, 4.6, 8.05

  3. 4.6, 2.95, 5.3, 12.75

  4. none of these

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Cr ${ 1s }^{ 2 }\quad { 2s }^{ 2 }\quad { 2p }^{ 6 }\quad { 3s }^{ 2 }\quad { 3p }^{ 6 }\quad { 3d }^{ 5 }\quad { 4s }^{ 1 }$

${ 24 }^{ - }$
Using later's rule:
for $3de^{ - }\quad { Z } _{ eff }=24-(4\times 0.35)-(18\times 1)=4.6$
for $4se^{ - }\quad { Z } _{ eff }=24-(0\times 0.35)-(13\times 0.85)-(10\times 1)=2.95$
$Cr^{ 3+ }\quad ({ 1s }^{ 2 }\quad { 2s }^{ 2 }\quad { 2p }^{ 6 }\quad { 3s }^{ 2 }\quad { 3p }^{ 6 }\quad { 3d }^{ 3 })$
for $3de^{ - }\quad { Z } _{ eff }=24-(2\times 0.85)-18=5.3$
for $3se^{ - }\quad { Z } _{ eff }=24-(7\times 0.35)-(8\times 0.85)-(2\times 1)=12.75$
                                         ${ s }^{ 1 }p^{ 6 }$           $2({ s }^{ 2 }p^{ 6 })$                    ${ 1s }^{ 2 }$

Multiple choice chemistry periodicity periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

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

  1. 2

  2. 1.7

  3. 1.85

  4. 1.65

Reveal answer Fill a bubble to check yourself
A Correct answer
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.


Multiple choice chemistry periodicity periodic trends in physical properties properties and trend trends in periodic table electronic configuration and valency electron configuration

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

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The screening effect (shielding) depends on the shape of the orbitals. s-orbitals are closest to the nucleus and provide the most shielding, while d-orbitals are more diffuse and have poor shielding ability.

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

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

  1. valency

  2. affinity

  3. bonding

  4. energy levels

Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

The valency of nitrogen in nitrogen dioxide is:

  1. 1

  2. 2

  3. 3

  4. 4

Reveal answer Fill a bubble to check yourself
D Correct answer
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 $. 

Multiple choice chemistry structure of the atom electronic configuration and valency electron configuration periodic trends in physical properties

Valency of magnesium and oxygen in $MgO$ is:

  1. one and one

  2. two and two

  3. one and two

  4. two and one

Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2$ .