MCQ
Azimuthal quantum number for last electron of $Na$ atom is
- A$1$
- B$2$
- C$3$
- ✓$0$
✓
Answer
Correct option: D.
$0$
d
(d) $N{a_{11}} = 1{s^2}2{s^2}{p^6}3{s^2}$
(d) $N{a_{11}} = 1{s^2}2{s^2}{p^6}3{s^2}$
$n = 3,\,l = 0,\,m = 0$ and $s = + \frac{1}{2}.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$Co|Co^{2+}(C_2)||Co^{2+}(C_1)|Co$, for this cell, $\Delta G$ is negative if
→After understanding the assertion and reason, choose the correct option.
Assertion : In the bonding molecular orbital $(MO)$ of $H_2,$ electron density is increased between the nuclei.
Reason : The bonding $MO$ is ${\psi _A}\, + \,\,{\psi _{B,}}$ which shows destructive interference of the combining electron waves.
→Assertion : In the bonding molecular orbital $(MO)$ of $H_2,$ electron density is increased between the nuclei.
Reason : The bonding $MO$ is ${\psi _A}\, + \,\,{\psi _{B,}}$ which shows destructive interference of the combining electron waves.
Aniline reacts with mixed acid $($conc. $HNO _3$ and conc. $H _2 SO _4 )$ at $288 K$ to give $P (51 \%)$,$Q (47 \%)$ and $R (2 \%)$. The major product$(s)$ of the following reaction sequence is $($are$)$
$\text { 1) } Ac _2 O \text {, pyridine } \text { 1) } Sn / HCl $
$R \xrightarrow{2) Br _2, CH _3 CO _2 H } S \xrightarrow{2) Br _2 / H _2 O \text { (excess) }} \text { major product(s) }$
$\text { 3) } H _3 O ^{+} \text {3) } NaNO _2, HCl / 273-278 K$
$\text { 4) } NaNO _2, HCl / 273-278 K \quad \text { 4) } H _3 PO _2$
$\text { 5) } EtOH , \Delta$
→$\text { 1) } Ac _2 O \text {, pyridine } \text { 1) } Sn / HCl $
$R \xrightarrow{2) Br _2, CH _3 CO _2 H } S \xrightarrow{2) Br _2 / H _2 O \text { (excess) }} \text { major product(s) }$
$\text { 3) } H _3 O ^{+} \text {3) } NaNO _2, HCl / 273-278 K$
$\text { 4) } NaNO _2, HCl / 273-278 K \quad \text { 4) } H _3 PO _2$
$\text { 5) } EtOH , \Delta$
The exothermic formation of $ClF_3$ is represented by the equation
→$C{l_{2\left( g \right)}} + 3{F_2}_{\left( g \right)} \rightleftharpoons 2C{l_3}_{\left( g \right)};\Delta {H_r} = - 329\,kJ$
Which of the following will increase the quantity of $ClF_3$ in an equilibrium mixture of $Cl_2, F_2$ and $ClF_3$ ?
While performing a thermodynamics experiment, a student made the following observations, $HCl + NaOH \rightarrow NaCl + H _{2} O \Delta H =-57.3\, kJ\,mol ^{-1}$
→$CH _{3} COOH + NaOH \rightarrow CH _{3} COONa + H _{2} O$ $(\Delta H =-55.3\,kJ\,mol ^{-1})$ The enthalpy of ionization of $CH _{3} COOH$ as calculated by the student is $kJ mol ^{-1}$. (nearest integer)
Which is mild oxidising agent
product $(A)$ is
→The oxidation number of $Cr$ in ${K_2}C{r_2}{O_7}$ is
→An aqueous solution of $NaCl$ shows the depression of freezing point of water equal to $0.372\,K$. The boiling point of $BaCl_2$ solution of same molality will be .........$^oC$ .$[K_f \,(H_2O)= 1.86\,K\,kg\,mol^{-1}$ ; $K_b\,(H_2O) = 0.52\,K\,kg\,mol^{-1} ]$
→$M{n^{ + 2 }}$ can be converted into $M{n^{7 + }}$ by reacting with
→