Tag: trends in periodic table

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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, 
Z 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.