Q: When copper is heated with \$conc. HNO_{3}\$ it produces :

When copper is heated with conc. \$\displaystyle HNO_3\$ it produces cupric nitrate \$\displaystyle Cu(NO_{3})_{2}\$ and nitrogen dioxide \$\displaystyle NO_{2}\$. \$\displaystyle Cu + 4 HNO_3 \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_2O\$Hence, option \$A\$ is correct.

Q: Among the oxides of nitrogen: \${N}_{2}{O}_{3}, {N}_{2}{O}_{4}\$ and \${N}_{2}{O}_{5}\$; the molecule(s) having nitrogen-nitrogen bond is/are:

Among the oxides of nitrogen: \${N}_{2}{O}_{3}, {N}_{2}{O}_{4}\$ and \${N}_{2}{O}_{5}\$; the molecule(s) having nitrogen-nitrogen bond are\${N}_{2}{O}_{3} \$ and \${N}_{2}{O}_{4}\$. Whereas, \${N}_{2}{O}_{5}\$ has nitrogen-oxygen bonds only.

Q: How many of the species are paramagnetic?\$N_2O,\ NO,\ NO_2,\ O_2,\ NO_2^+,\ NO^+,\ CO^+\$

Number of valance electrons in \$N_2O= 5+6+5 = 16\$All electrons can be paired up and there is no free electron. \$N_2O\$ molecule is therefore diamagnetic. Number of valance electrons in \$NO= 5+6 = 11\$All electrons can not be paired up and there is an odd electron responsible for magnetism. \$NO\$ molecule is therefore paramagnetic. Number of valance electrons in \$NO_2= 5+6+6 = 17\$All electrons can not be paired up and there is an odd electron responsible for magnetism. \$NO_2\$ molecule is therefore paramagnetic. Number of valance electrons in \$O_2= 6+6 = 12\$All electrons can be paired up and there is no free electron. \$O_2\$ molecule is exception and is paramagnetic.Number of valance electrons in \$NO_2^+= 5+6+6-1 = 16\$All electrons can be paired up and there is no free electron. \$NO_2^+\$ molecule is therefore diamagnetic.Number of valance electrons in \$NO^+= 5+6-1 = 10\$All electrons can be paired up and there is no free electron. \$NO^+\$ molecule is therefore diamagnetic.Number of valance electrons in \$CO^+= 4+6-1= 9\$All electrons can not be paired up and there is an odd electron. \$CO^+\$ molecule is therefore paramagnetic.Overall 4 species are paramagnetic including \$NO,\ NO_2,\ O_2\$ and \$CO^+\$

Q: Which of the following is the correct electron dot structure of \$N_2O\$ molecule?

The electron dot structure is \${ N }_{ 2 }O\$

Q: A brown ring is formed in the ring test for \$NO_3\$ ion. It is due to the formation of:

The identification and conformation test of nitrate ion ( \$NO_3^{–}\$) using conc. \$H_2SO_4 and freshly prepared FeSO4 , is called as 'Brown Ring Test. In a test tube take the \$NO_3^{–}\$ containing sample .Then add conc.\$H_2SO_4\$ dropwise with gentle shaking and then heat it shortly using Bunsen burner & add freshly prepared greenish \$FeSO_4\$. Then, we can observe a brown ring formed in that test tube due to the formation of brown colored \$[Fe(H_2O)5 (NO)]SO_4 complex.Hence option A is correct.

Q: Out of \$NO, NO_2, N_2O, N_2O_4\$ and \$N_2O_2\$, the paramagnetic species are__________.

\$NO, NO_2\$ both have one unpaired electron so they both are paramagnetic in nature i.e. is odd electron system.Other species do not have unpaired electron so they are diamagnetic in nature.

Q: Most acidic oxide among the following is :

\$\overset{+5}{N_{2}O_{5}}\$ As oxidation state increases acidic nature increases. Acidic nature decreases down the group

Q: Which oxide is colourless and neutral ?

\$N_2O_3\$ is a blue solid but other oxides are colourless. Among the others only \$N_2O\$ (nitrous oxide, also called laughing gas) Is neutral to litmus. Note: Acidic character of oxides generally increases as the number of oxygen atom increases.

Q: Why does \$N{O}_{2}\$ dimerise?

The dimerisation of \$\displaystyle NO_2\$ is due to presence of unpaired electron on nitrogen.

Question 1

When copper is heated with $conc. HNO_{3}$ it produces :

Accepted Answer

When copper is heated with conc. $\displaystyle HNO_3$ it produces cupric nitrate  $\displaystyle Cu(NO_{3})_{2}$ and nitrogen dioxide $\displaystyle NO_{2}$. $\displaystyle Cu + 4 HNO_3 \rightarrow Cu(NO_{3})_{2} + 2NO_{2} + 2H_2O$Hence, option $A$ is correct.

Question 2

Among the oxides of nitrogen: ${N}_{2}{O}_{3}, {N}_{2}{O}_{4}$ and ${N}_{2}{O}_{5}$; the molecule(s) having nitrogen-nitrogen bond is/are:

Accepted Answer

Among the oxides of nitrogen: ${N}_{2}{O}_{3}, {N}_{2}{O}_{4}$ and ${N}_{2}{O}_{5}$; the molecule(s) having nitrogen-nitrogen bond are${N}_{2}{O}_{3} $ and ${N}_{2}{O}_{4}$. Whereas, ${N}_{2}{O}_{5}$ has nitrogen-oxygen bonds only.

Question 3

Select the incorrect statement(s) about $N_{2} F_{4}$ and $N_{2} H_{4}$ .
(i) In $N_{2} F_{4}$ , d-orbitals are contracted by electronegative fluorine atoms, but d-orbital contraction is not possible by H-atoms in $N_{2} H_{4}$ .
(ii) The $N - N$ bond energy in $N_{2} F_{4}$ is more than $N - N$ bond energy in $N_{2} H_{4}$ .
(iii) The $N - N$ bond length in $N_{2} F_{4}$ is more than that in $N_{2} H_{4}$ .
(iv) The $N - N$ bond length in $N_{2} F_{4}$ is less than that in $N_{2} H_{4}$ .
Choose the correct code.

Accepted Answer

1) s-character tend to concentrate in orbitals that are directed toward electropositive groups and p-character tends to concentrate in orbitals that are directed toward electronegative groups. 2) It can be explained on the basis of Bent's rule. In $N_{2}F_{4}$, $N-N$ bond has more s-character, hence, bond length decreases. While in $N_{2}H_{4}$, $N-N$ bond has less s-character (i.e. more p-character), hence bond length increases.

Question 4

How many of the species are paramagnetic?$N_2O,\ NO,\ NO_2,\  O_2,\  NO_2^+,\  NO^+,\ CO^+$

Accepted Answer

Number of valance electrons in $N_2O= 5+6+5 = 16$All electrons can be paired up and there is no free electron. $N_2O$ molecule is therefore diamagnetic. Number of valance electrons in $NO= 5+6 = 11$All electrons can not be paired up and there is an odd electron responsible for magnetism. $NO$ molecule is therefore paramagnetic. Number of valance electrons in $NO_2= 5+6+6 = 17$All electrons can not be paired up and there is an odd electron responsible for magnetism. $NO_2$ molecule is therefore paramagnetic. Number of valance electrons in $O_2= 6+6 = 12$All electrons can be paired up and there is no free electron. $O_2$ molecule is exception and is paramagnetic.Number of valance electrons in $NO_2^+= 5+6+6-1 = 16$All electrons can be paired up and there is no free electron. $NO_2^+$ molecule is therefore diamagnetic.Number of valance electrons in $NO^+= 5+6-1 = 10$All electrons can be paired up and there is no free electron. $NO^+$ molecule is therefore diamagnetic.Number of valance electrons in $CO^+= 4+6-1= 9$All electrons can not be paired up and there is an odd electron. $CO^+$ molecule is therefore paramagnetic.Overall 4 species are paramagnetic including $NO,\ NO_2,\ O_2$ and $CO^+$

Question 5

Which of the following is the correct electron dot structure of $N_2O$ molecule?

Accepted Answer

The electron dot structure is ${ N }_{ 2 }O$

Question 6

A brown ring is formed in the ring test for $NO_3$ ion. It is due to the formation of:

Accepted Answer

&#10;&#10;The identification and conformation test of nitrate ion  ( $NO_3^{&#8211;}$) using conc. $H_2SO_4 and freshly&#10;prepared FeSO4 , is called  as 'Brown&#10;Ring Test. In a test tube take the $NO_3^{&#8211;}$  containing sample .Then  add conc.$H_2SO_4$ dropwise with gentle&#10;shaking and then heat it shortly using Bunsen burner & add freshly prepared&#10;greenish $FeSO_4$. Then, we can observe &#10;a brown ring formed in that test tube due to the formation of brown colored&#10;$[Fe(H_2O)5 (NO)]SO_4  complex.Hence option A is correct.&#10;&#10;

Question 7

Out of $NO, NO_2, N_2O, N_2O_4$ and $N_2O_2$, the paramagnetic species are__________.

Accepted Answer

$NO, NO_2$ both have one unpaired electron so they both are paramagnetic in nature i.e. is odd electron system.Other species do not have unpaired electron so they are diamagnetic in nature.

Question 8

Most acidic oxide among the following is :

Accepted Answer

$\overset{+5}{N_{2}O_{5}}$ As oxidation state increases acidic nature increases. Acidic nature decreases down the group

Question 9

Which oxide is colourless and neutral ?

Accepted Answer

$N_2O_3$ is a blue solid but other oxides are colourless. Among the others only $N_2O$ (nitrous oxide, also called laughing gas) Is neutral to litmus. Note: Acidic character of oxides generally increases as the number of oxygen atom increases.

Question 10

Why does $N{O}_{2}$ dimerise?

Accepted Answer

The dimerisation of $\displaystyle  NO_2$ is due to presence of unpaired electron on nitrogen.

Oxides of nitrogen

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