In $$XeF_2\cdot2SbF_5$$

There are two Xe-F bonds with bond lengths $$184$$ pm and $$235$$ pm

F is forming bridge between Xe and Sb

Both $$1)$$ and $$2)$$ are correct

Neither $$1)$$ or $$2)$$ is correct

MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 5

Mono atomic element among the following is:

0%
phosphorus
0%
oxygen
0%
krypton
0%
sulphur

The inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total

$p-$ and

$s-$ electrons, is:

0%
$He$
0%
$Ne$
0%
$Ar$
0%
$Kr$

Explanation

$He, Ne$ and

$Ar$ do not have any

$d$ electrons, while the difference of their

$s$ and

$p$ electrons is not zero.

The electronic configuration of krypton (atomic number is 36) is:

$1s^2,2s^2,2p^6,3s^2,3p^6,3d^{10},4s^2,4p^6$ .

Total

$s-$ electrons are 8 and total

$p-$ electrons are 18.

Thus, the difference is 10 and total d-electrons are 10 which are equal.

So, option

$D$ is correct answer.

Which of the following sets of atomic numbers belongs to chalcogens?

0%
8, 52, 84
0%
9, 17, 35
0%
20, 37, 56
0%
14, 32, 50

Explanation

$\underset{VI group elements so belongs to chalcogens}{\underbrace{ 8:oxygen 52:Tellurium 84:Polonium}}$

According to Coulson the high reactivity of fluorine than other halogens is because of its:

0%
high electronegativity
0%
lesser internuclear repulsions
0%
greater repulsions between non bonding electrons
0%
All the above

Elements with atomic numbers 9, 17, 35, 53 are collectively known as:

0%
chalcogens
0%
halogens
0%
lanthanides
0%
rare gases

Explanation

Elements having the atomic number:

9 is fluorine:

$(He)2s^22p^5$
17 is chlorine:

$(Ne)3s^23p^5$
35 is bromine:

$(Ar)4s^24p^5$
53 is iodine:

$(Kr)5s^25p^5$

We can observe here that all these elements having the general electronic configuration of

$ns^2np^5$ , hence these elements belong to

$17^{th}$ group elements.

The inert gas present in the second long period is:

0%
$Kr$
0%
$Xe$
0%
$Ar$
0%
$Rn$

Explanation

Periods

$4$ and

$5$ are called long periods as they contain

$18$ elements each.

$4$ th period is called the first long period and the inert gas in this period is

$Kr$ .

$5$ th period is the second long period where

$Xe$ is the inert gas element.

The outer most electronic configuration of the most electronegative element is:

0%
$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{3}$
0%
$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{4}$
0%
$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{5}$
0%
$\mathrm{n}\mathrm{s}^{2}\mathrm{n}\mathrm{p}^{6}$

Explanation

Halogens are most electronegative.

Therefore outer electronic configuration is

$ns^{2}\ np^{5}.$

Option C is correct.

Which of the following does not contain fluorine?

0%
Fluorspar
0%
Cryolite
0%
Flouorospatite
0%
Chile saltpeter

Explanation

Chile saltpeter -

$NaNO_{3}$ does not contain Fluorine.
All contain Fluorine

$\left\{\begin{matrix} Fluorspar - CaF_{2}\\Cryolite - Na_{3}AlF_{6} \\Flourospatite - Ca_{5}(PO_{4})_{3}F \end{matrix}\right.$

Which of the following exist?

0%
$ICl_{3}$
0%
$ClF_{3}$
0%
$IF_{7}$
0%
$BrF_{3}$

Explanation

In interhalogen compounds, take example

$XX'_n$ , the central atom is

$X$ here. The compound exists only when

$X$ is of larger size and

$X'$ is of smaller size and

$X$ is more electronegative than

$X'$ . From this condition,

$ICl_3, \ IF_7$ and

$BrF_3$ exists.

$ClF_3$ does not exist because the difference between the size of

$Cl$ and

$F$ is very small.

Fluorine is highly reactive because:

0%
F-F bond energy is high
0%
F-F bond energy is low
0%
it is gaseous at room temperature
0%
F has a smaller size

Explanation

$F_{2}$ is highly reactive because of its low bond energy. Also, it has the highest electronegativity.

Which of the following statements is not correct?

0%
All halogens except flourine from OXO-acids
0%
Hypohalous acids HOCI,HOBr and HOI are all strong acids
0%
Hypochlorous acid is the most stable among hypohalous acids
0%
Chlorous acid is the only halous acid known

Explanation

There are also other halous acids like:

$HOBr_2$ which is Hydrobromous acid.
So D is wrong

The lightest gas which is non-inflammable is:

0%
${H}_{2}$
0%
$He$
0%
${N}_{2}$
0%
${A}$ r

Explanation

$H_2$ is the lightest gas but it is inflammable.

$He$ is the next lightest gas and it is non-inflammable.

The atomic number of astatine which belongs to the sixth period is:

0%
53
0%
54
0%
85
0%
86

$Al+X\overset{\Delta}{\rightarrow} Y \overset{H_{2}O}{\rightarrow} Z+NH_{3}$
Here

$X, Y$ and

$Z$ are respectively:

0%
$O_{2}, \; Al_{2}O_{3}, \; Al(OH)_{3}$
0%
$N_{2}, \; AlN, \; Al_{2}O_{3}$
0%
$N_{2}, \; AlN, \; Al(OH)_{3}$
0%
$CO_{2}, \; Al_{4}C_{3}, \; CH_{4}$

Explanation

$2Al+N_{2}\overset{\Delta}{\rightarrow}2AlN\overset{H_{2}O}{\rightarrow}Al(OH)_{3}+NH_{3}$

The inert gas which cannot be adsorbed over activated coconut charcoal?

0%
Ne
0%
Ar
0%
He
0%
Kr

Which of the following halogens does not form oxyacids at room temperature?

0%
$\mathrm{F}_{2}$
0%
$\mathrm{Cl}_{2}$
0%
$\mathrm{B}\mathrm{r}_{2}$
0%
$I_{2}$

Explanation

Fluorine does not hove d-sub shell so it can only have one oxidation state. Therefore, it cann't form oxy acids. Fluorine is the most electronegative element and always show

$-1$ oxiadation state. In oxyacids. reast of the halogens have positive oxidation state.

A degenerate gas is :

0%
He
0%
$\mathrm{O}_{2}$
0%
He(ll)
0%
$\mathrm{H}\mathrm{e}_{2}$

Explanation

$He(II)$ is a degenerate gas.
A degenerate gas is one where quantum mechanical effects on the electrons dominate the behavior.A degenerate electron gas can permanently withstand the self-gravitation. its energy cannot be radiated away.

The atomic weight of noble gases is obtained by using the relationship :

0%
Atomic weight = equivalent weight X valency
0%
Atomic weight = equivalent weight/valency
0%
At. weight = valency/equivalent weight
0%
2 x V.D. = molecular weight = atomic weight

Explanation

For Noble gases

$\underset{weight}{Molecular} = \underset{weight}{Atomic}$

$\underset{Density}{Vapour} = \frac{Atomic weight}{2}$

$\therefore 2\times U.D = molecular weight$

$= atomic weight$

To prepare

$XeF_6$ at 573 K, 50-60 atm pressure, the ratio of xenon and fluorine used is :

0%
1 : 2
0%
1 : 5
0%
1 : 20
0%
10 : 1

Explanation

The ratio of xenon and fluorine is 1:20 at 573K, 50-60 atm pressure to prepare

$XeF_{6}$ .

The correct order of enthalpy of vaporisation of noble gases is :

0%
$He < Ne < Ar < Kr < Xe$
0%
$He > Ne > Ar > Kr > Xe$
0%
$He < Ar < Kr < Ne < Xe$
0%
$He < Xe < Ar < Ne < Kr$

Explanation

Enthalpy of vaporization

$\propto$ Van der Waals forces

$\propto$ Molar mass

Thus, as molar mass increases, the enthalpy of vaporization increases.
Thus correct order of enthalpy of vapourisation of noble gases.

$Xe>Kr>Ar>Ne>He$

Assertion (A): Balloons made by nylon films are better for containing helium than the conventional rubber balloons.
Reason (R): R.M.S. velocity of helium is very high. So helium atoms can effuse out through rubber balloons.

0%
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$ .
0%
Both $A$ and $R$ are correct and $R$ is not the correct explanation of $A$ .
0%
$A$ is true but $R$ is false.
0%
$A$ is false but $R$ is true.

Explanation

Root mean square velocity of helium gas is high due to its low mass.

$RMS velocity \times \frac{1}{mass}$
Therefore, nylon films are better for containing helium than the conventional rubber balloons.

Compounds formed when the noble gases get entrapped in the cavities of crystal lattices of certain organic and inorganic compounds are known as :

0%
Interstitial compounds
0%
Clathrates
0%
Hydrates
0%
Picrates

The number of steps involved in the manufacturing of nitric acid through Ostwald's process is:

0%
1
0%
5
0%
3
0%
6

Explanation

Ostwald process to manufacture nitric acid-

Step

$1$ : Catalytic oxidation of Ammonia

A mixture of dry air and dry ammonia in the ratio of

$10 : 1$ by volume is compressed and then passed into a platinum gauze which acts as catalyst at about

$800 ℃$ .

$4 N{H}_{3} + 5 {O}_{2} \longrightarrow 4 NO + 6 {H}_{2}O + \text{Heat}$

Step

$2$ : Oxidation of nitric oxide

Nitric oxide combines with oxygen to form nitrogen dioxide at about 50°C.

$2 NO + {O}_{2} \longrightarrow 2 N{O}_{2}$

Step

$3$ : Absorption of nitrogen dioxide in water

The nitrogen dioxide and oxygen present in the air react with water to form nitric acid.

$4 N{O}_{2} + 2 {H}_{2}O + {O}_{2} \longrightarrow \underset{\text{Nitric acid}}{4 HN{O}_{3}}$

Nitric acid obtained is concentrated above

$50 \%$ . On further distillation,

$68 \%$ nitric acid is produced.

Hence, the number of steps involved in the manufacturing of nitric acid is 3.

The correct statement (s) regarding,

$\left( i \right)\,\,HClO\,\,\left( {ii} \right)\,\,\,HCl{O_2}\,\,\,\,\left( {iii} \right)\,\,\,HCl{O_3}$ and

$\left( {iv} \right)\,\,\,HCl{O_4}$ , is:

0%
The number of $Cl=O$ bond in $(ii)$ and $(iii)$ together is two.
0%
The numbers of lone pairs of electrons on $Cl$ in $(ii)$ and $(iii)$ together is three.
0%
The hybridization of $Cl$ in $(iv)$ is $s{p^3}$ .
0%
Amongst $(i)$ to $(iv)$ , the strongest acid is $(i)$ .

Which type of interaction is involved in the solubility of noble gas in water?

0%
Instantaneous dipole induced dipole interacation
0%
Dipole-dipole ineteractions
0%
Dipole-induced dipole interaction
0%
Ion-diplole interactions

Which of the following species is most unstable?

0%
$Li^-$
0%
$N^-$
0%
$C^-$
0%
$O^-$

Which of the following halogen form only one oxyacids (HOX)?

0%
$F$
0%
$Br$
0%
$I$
0%
$Cl$

Which of the following noble gases can be called the hidden one?

0%
$Xe$
0%
$He$
0%
$Ar$
0%
$Kr$

${\text{Xe}}{{\text{F}}_{\text{6}}}$ on partial hydrolysis with water produces a compound 'X'. The same compound 'X' is formed when

${\text{Xe}}{{\text{F}}_{\text{6}}}$ reacts with silica. The compound 'X' is:

0%
${\text{Xe}}{{\text{F}}_2}$
0%
$\,\,{\text{Xe}}{{\text{F}}_4}$
0%
$\,\,{\text{XeO}}{{\text{F}}_4}$
0%
$\,\,{\text{Xe}}{{\text{O}}_3}$

In which of the following pairs, the second atom is larger than the first?

0%
$N,\ P$
0%
$Br,\ Cl$
0%
$Ba,\ Sr$
0%
$Mg,\ Al$

Which of the following substance produces oxygen on heating?

0%
Potassium Permanganate
0%
Potassium Iodide
0%
Potassium chloride
0%
Potassium hydroxide

In the manufacture of nitric acid by Ostwald's process, ___________ as a dehydrating agent is used to convert 68% by mass of

$HNO_3$ to 98% by mass of

$HNO_3$ .

0%
$H_2SO_4$
0%
Anhydrous $CaCl_2$
0%
Hot $Al_2O_3$
0%
None of these

Which of the following compound does not liberate ammonia gas on refluxing with aqueous NaOH?

A nitrate which on heating liberates oxygen as the only gaseous product is:

0%
${ NaNO }_{ 3 }$
0%
$Mg{ \left( NO_{ 3 } \right) }_{ 2 }$
0%
${ AgNO }_{ 3 }$
0%
$Pb\left( { NO }_{ 3 } \right) _{ 2 }$

Which of the following has highest chlorine content ?

0%
Pyrene
0%
DDT
0%
Chloral
0%
Gammaxene

To dry ammonia gas the drying agent used is:

0%
anhydrous $CaCl_2$
0%
concentrated $H_2SO_4$
0%
$P_4O_{10}$
0%
$CaO$

The nomenclature of

$ICl$ is iodine chloride because:

0%
Size of $I$ < Size of $Cl$
0%
Atomic number of $I$ > Atomic number of $Cl$
0%
E.N. of $I$ < E.N. of $Cl$
0%
E.A of $I$ < E. A. of $Cl$

Which of the following is a tetratomic solid?

0%
Phosphorous
0%
Nitrogen
0%
Potassium
0%
Calcium

Which of the following compound on heating does not give ammonia?

0%
$( \mathrm { NH }_4 )_2 \mathrm { SO_4 }$
0%
$( \mathrm { NH }_4 )_2 \mathrm { CO_3 }$
0%
$( \mathrm { NH }_4 \mathrm { NO_3 }$
0%
$NH_4Cl$

Ammonia can be obtained by adding water to:

0%
ammonium chloride
0%
ammonium nitrite
0%
magnesium nitride
0%
magnesium nitrate

Natural gas is a major feed stock for the production of ________

0%
Ammonia
0%
potassium
0%
sodium
0%
phosphorus

The non-existent species is

0%
$XeF_4$
0%
$BrF_4$
0%
$SbF_5$
0%
$PF_5$

Both N(Si

${ H }_{ 3 })_{ 3 }$ and

$NH(SiH_{ 3 })_{ 2 }$ compounds have trigonal planar skeleton. Incorrect statement about both compounds is:

0%
SiNSi bond angle in $NH(SiH_{ 3 })_{ 2 }$ >SiNSi bond angle in $N(SiH_{ 3 })_{ 3 }$
0%
N-Si bond length in $NH(SiH_{ 3 })_{ 2 }$ > N-Si bond length in $N(SiH_{ 3 })_{ 3 }$
0%
N-Si bond length in $NH(SiH_{ 3 })_{ 2 }$ < N-Si bond length in n $N(SiH_{ 3 })_{ 3 }$
0%
Back bonding strength in $NH(SiH_{ 3 })_{ 2 }$ > Back bonding strength in $N(SiH_{ 3 })_{ 3 }$

White and yellow phosphorous are:

0%
Allotropes
0%
Isomers
0%
Isobars
0%
Isotones

Most energetic species among the following is:

0%
$H _ { 2 }$
0%
$Ne$
0%
$F$
0%
$F _ { 2 }$

Greater reactivity of

$F_2$ is not due to____

0%
low energy of F-F bond
0%
small size
0%
high heat of hydration
0%
high energy of F-F bond

White and yellow phosphorus are -

0%
Allotropes
0%
Isomoers
0%
Isobars
0%
Isotones

$XeF_2\cdot2SbF_5$

0%
F is forming bridge between Xe and Sb
0%
There are two Xe-F bonds with bond lengths $184$ pm and $235$ pm
0%
Both $1)$ and $2)$ are correct
0%
Neither $1)$ or $2)$ is correct

White phosphorus on reaction with concentrated

$NaOH$ solution in an inert atmosphere of

${CO}_{2}$ gives phosphine and compound

$(X)$ .

$(X)$ on acidification with

$HCl$ gives compound

$(Y)$ . The basicity of compound

$(Y)$ is :

0%
$4$
0%
$3$
0%
$1$
0%
$2$

Explanation

The basicity of compound

$Y$ is 1.

Hence, option

$C$ is correct.

1476527_1697402_ans_e210dabbeb3e4cee9e3e74c0af888231.png

Very pure

$N_{2}$ can be obtained by:

0%
Thermal decomposition of ammonium dichromate
0%
Treating aqueous solution of $NH_{4}Cl$ and $NaNO_{2}$
0%
Liquifaction and fractional distillation of liquid air
0%
Thermal decomposition of sodium azide

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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