Classification of Elements and Periodicity in Properties
89532
Among $\mathrm{Me}_3 \mathrm{~N}_{,} \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}$ and $\mathrm{MeCN}(\mathrm{Me}=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order
In $\mathrm{Me}_3 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^3$ hybrid In $\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^2$ hybrid. In $\mathrm{MeCN} \Rightarrow \mathrm{N}$ is sp-hybrid Therefore, the electronegativity of element depend on $\mathrm{s}$-character, s-character is increase then electronegativity increases. $$ \text { MeCN }>\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}>\mathrm{Me}_3 \mathrm{~N} $$
WB-JEE-2017
Classification of Elements and Periodicity in Properties
89533
The correct order of electronegativities of $\mathrm{N}, \mathrm{O}$, $F$ and $P$ is
1 $\mathrm{F}>\mathrm{O}>\mathrm{P}>\mathrm{N}$
2 F $>$ O $>\mathrm{N}>\mathrm{P}$
3 $\mathrm{N}>$ O $>$ F $>$ P
4 F $>$ N $>$ P $>$ O
Explanation:
In a period, electronegativity increases from left to right. $$ \mathrm{F}>\mathrm{O}>\mathrm{N} $$ In a group, electronegativity decreases down the group. Hence, $\mathrm{N}>\mathrm{P}$ So, the order of electronegativity are- $$ \begin{array}{llll} \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P} \\ 4 & 3.5 & 3 & 2.19 \end{array} $$
Karnataka-CET-2012
Classification of Elements and Periodicity in Properties
89527
Assertion: $\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation, but $\mathrm{F}$ doesn't. Reason: This is because it is the most electronegative element of the group.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation but fluorine $(F)$ does not show disproportionation tendency since it can only take up electron (oxidation state $=-1$ ) and cannot lose electron. This is because it is the most electronegative element of the group.
AIIMS 25 May 2019 (Evening)
Classification of Elements and Periodicity in Properties
89547
Let electronegativity, ionization energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?
1 $\mathrm{EN}=\mathrm{IP} \times \mathrm{EA}$
2 $\mathrm{EN}=\frac{\mathrm{IP}}{\mathrm{EA}}$
3 $\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$
4 $\mathrm{EN}=\mathrm{IP}-\mathrm{EA}$
Explanation:
Given, Electronegativity $=\mathrm{EN}$ Ionisation energy $=$ IP Electron affinity $=\mathrm{EA}$ Mulliken has defined electronegativity as the average of ionization potential and electronegativity. $$ \mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2} $$
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Classification of Elements and Periodicity in Properties
89532
Among $\mathrm{Me}_3 \mathrm{~N}_{,} \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}$ and $\mathrm{MeCN}(\mathrm{Me}=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order
In $\mathrm{Me}_3 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^3$ hybrid In $\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^2$ hybrid. In $\mathrm{MeCN} \Rightarrow \mathrm{N}$ is sp-hybrid Therefore, the electronegativity of element depend on $\mathrm{s}$-character, s-character is increase then electronegativity increases. $$ \text { MeCN }>\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}>\mathrm{Me}_3 \mathrm{~N} $$
WB-JEE-2017
Classification of Elements and Periodicity in Properties
89533
The correct order of electronegativities of $\mathrm{N}, \mathrm{O}$, $F$ and $P$ is
1 $\mathrm{F}>\mathrm{O}>\mathrm{P}>\mathrm{N}$
2 F $>$ O $>\mathrm{N}>\mathrm{P}$
3 $\mathrm{N}>$ O $>$ F $>$ P
4 F $>$ N $>$ P $>$ O
Explanation:
In a period, electronegativity increases from left to right. $$ \mathrm{F}>\mathrm{O}>\mathrm{N} $$ In a group, electronegativity decreases down the group. Hence, $\mathrm{N}>\mathrm{P}$ So, the order of electronegativity are- $$ \begin{array}{llll} \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P} \\ 4 & 3.5 & 3 & 2.19 \end{array} $$
Karnataka-CET-2012
Classification of Elements and Periodicity in Properties
89527
Assertion: $\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation, but $\mathrm{F}$ doesn't. Reason: This is because it is the most electronegative element of the group.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation but fluorine $(F)$ does not show disproportionation tendency since it can only take up electron (oxidation state $=-1$ ) and cannot lose electron. This is because it is the most electronegative element of the group.
AIIMS 25 May 2019 (Evening)
Classification of Elements and Periodicity in Properties
89547
Let electronegativity, ionization energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?
1 $\mathrm{EN}=\mathrm{IP} \times \mathrm{EA}$
2 $\mathrm{EN}=\frac{\mathrm{IP}}{\mathrm{EA}}$
3 $\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$
4 $\mathrm{EN}=\mathrm{IP}-\mathrm{EA}$
Explanation:
Given, Electronegativity $=\mathrm{EN}$ Ionisation energy $=$ IP Electron affinity $=\mathrm{EA}$ Mulliken has defined electronegativity as the average of ionization potential and electronegativity. $$ \mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2} $$
Classification of Elements and Periodicity in Properties
89532
Among $\mathrm{Me}_3 \mathrm{~N}_{,} \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}$ and $\mathrm{MeCN}(\mathrm{Me}=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order
In $\mathrm{Me}_3 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^3$ hybrid In $\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^2$ hybrid. In $\mathrm{MeCN} \Rightarrow \mathrm{N}$ is sp-hybrid Therefore, the electronegativity of element depend on $\mathrm{s}$-character, s-character is increase then electronegativity increases. $$ \text { MeCN }>\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}>\mathrm{Me}_3 \mathrm{~N} $$
WB-JEE-2017
Classification of Elements and Periodicity in Properties
89533
The correct order of electronegativities of $\mathrm{N}, \mathrm{O}$, $F$ and $P$ is
1 $\mathrm{F}>\mathrm{O}>\mathrm{P}>\mathrm{N}$
2 F $>$ O $>\mathrm{N}>\mathrm{P}$
3 $\mathrm{N}>$ O $>$ F $>$ P
4 F $>$ N $>$ P $>$ O
Explanation:
In a period, electronegativity increases from left to right. $$ \mathrm{F}>\mathrm{O}>\mathrm{N} $$ In a group, electronegativity decreases down the group. Hence, $\mathrm{N}>\mathrm{P}$ So, the order of electronegativity are- $$ \begin{array}{llll} \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P} \\ 4 & 3.5 & 3 & 2.19 \end{array} $$
Karnataka-CET-2012
Classification of Elements and Periodicity in Properties
89527
Assertion: $\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation, but $\mathrm{F}$ doesn't. Reason: This is because it is the most electronegative element of the group.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation but fluorine $(F)$ does not show disproportionation tendency since it can only take up electron (oxidation state $=-1$ ) and cannot lose electron. This is because it is the most electronegative element of the group.
AIIMS 25 May 2019 (Evening)
Classification of Elements and Periodicity in Properties
89547
Let electronegativity, ionization energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?
1 $\mathrm{EN}=\mathrm{IP} \times \mathrm{EA}$
2 $\mathrm{EN}=\frac{\mathrm{IP}}{\mathrm{EA}}$
3 $\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$
4 $\mathrm{EN}=\mathrm{IP}-\mathrm{EA}$
Explanation:
Given, Electronegativity $=\mathrm{EN}$ Ionisation energy $=$ IP Electron affinity $=\mathrm{EA}$ Mulliken has defined electronegativity as the average of ionization potential and electronegativity. $$ \mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2} $$
Classification of Elements and Periodicity in Properties
89532
Among $\mathrm{Me}_3 \mathrm{~N}_{,} \mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}$ and $\mathrm{MeCN}(\mathrm{Me}=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order
In $\mathrm{Me}_3 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^3$ hybrid In $\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N} \Rightarrow \mathrm{N}$ is $\mathrm{sp}^2$ hybrid. In $\mathrm{MeCN} \Rightarrow \mathrm{N}$ is sp-hybrid Therefore, the electronegativity of element depend on $\mathrm{s}$-character, s-character is increase then electronegativity increases. $$ \text { MeCN }>\mathrm{C}_5 \mathrm{H}_5 \mathrm{~N}>\mathrm{Me}_3 \mathrm{~N} $$
WB-JEE-2017
Classification of Elements and Periodicity in Properties
89533
The correct order of electronegativities of $\mathrm{N}, \mathrm{O}$, $F$ and $P$ is
1 $\mathrm{F}>\mathrm{O}>\mathrm{P}>\mathrm{N}$
2 F $>$ O $>\mathrm{N}>\mathrm{P}$
3 $\mathrm{N}>$ O $>$ F $>$ P
4 F $>$ N $>$ P $>$ O
Explanation:
In a period, electronegativity increases from left to right. $$ \mathrm{F}>\mathrm{O}>\mathrm{N} $$ In a group, electronegativity decreases down the group. Hence, $\mathrm{N}>\mathrm{P}$ So, the order of electronegativity are- $$ \begin{array}{llll} \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{P} \\ 4 & 3.5 & 3 & 2.19 \end{array} $$
Karnataka-CET-2012
Classification of Elements and Periodicity in Properties
89527
Assertion: $\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation, but $\mathrm{F}$ doesn't. Reason: This is because it is the most electronegative element of the group.
1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{Cl}, \mathrm{Br}$ and I show disproportionation but fluorine $(F)$ does not show disproportionation tendency since it can only take up electron (oxidation state $=-1$ ) and cannot lose electron. This is because it is the most electronegative element of the group.
AIIMS 25 May 2019 (Evening)
Classification of Elements and Periodicity in Properties
89547
Let electronegativity, ionization energy and electron affinity be represented as EN, IP and EA respectively. Which one of the following equation is correct according to Mulliken?
1 $\mathrm{EN}=\mathrm{IP} \times \mathrm{EA}$
2 $\mathrm{EN}=\frac{\mathrm{IP}}{\mathrm{EA}}$
3 $\mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2}$
4 $\mathrm{EN}=\mathrm{IP}-\mathrm{EA}$
Explanation:
Given, Electronegativity $=\mathrm{EN}$ Ionisation energy $=$ IP Electron affinity $=\mathrm{EA}$ Mulliken has defined electronegativity as the average of ionization potential and electronegativity. $$ \mathrm{EN}=\frac{\mathrm{IP}+\mathrm{EA}}{2} $$
