Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

The decomposition of phosphine (PH_(3)) on tungsten at low pressure is a first order reaction. It is because the

Answer»

SOLUTION :`(t_(1))/(t_(2))=((a_(2))/(a_(1)))^(n-1)`
`a_(1)="290 MM of Hg "t_(1)="263 seconds"`
`a_(2)="360 mm of Hg "t_(2)="212 seconds"`
`(263)/(212)=((360)/(290))^(n-1)`
`1.24=(1.24)^(n-1)`
`n-1=1`
`n=1+1=2`
Hence, the REACTION is of second order.
Discussion
2.

In the tet of NO_(3)^(-)ion, the dark brown ring complex [Fe(H_(2)O)_(5)(NO)]SO_(4) is formed which of the following is false for this complex?

Answer»

The brown ring complex is FORMED between nitric oxide (formed as a result to reduction of the nitrate ION by the `Fe^(2+)` ions) and `Fe^(2+)` ions
Iron and `NO` both have `+1` charge
The complex species can be represented as `[Fe^(+)(H_(2)O)_(5)(NO)]^(2+)`
Iron has `+2` oxidation STATE and `NO` is neutral

Solution :NA
Discussion
3.

In the test for iodine, I_(2) is treated with sodium thiosulphate (Na_(2)S_(2)O_(3)) : Na_(2)S_(2)O_(3)+I_(2) to NaI +..

Answer»

`Na_(2)S_(4)O_(6)`
`Na_(2)SO_(4)`
`Na_(2)S`
`Na_(3)ISO_(4)`

ANSWER :A
Discussion
4.

In the talling of mercury ozone oxidies X to Y, X and Y are respectively

Answer»

`Hg, Hg(I)O`
`Hg, Hg(II)O`
`Hg(I)O, Hg(II)O`
`Hg(II)O, Hg(I)O`

SOLUTION :`2HG + O_(3) to Hg_(2)O + O_(2)`
Discussion
5.

In the tailing of mercury ozone oxidises X to Y.X and Y are respectively

Answer»

HG,Hg(I)O
Hg,Hg(II)O
Hg(I)O,Hg(II)O
Hg(II)O,Hg(I)O

Answer :A
Discussion
6.

In the tailing of mercury ozone oxidises X to Y, X and Y are respectively

Answer»

`HG, Hg (I)O`
`Hg,Hg(II)O`
`Hg(I)O, Hg (II)O`
`Hg(II)O, Hg(I)O`

ANSWER :A
Discussion
7.

In thesyudy of cholrinatio of propane, fourproducts (A,B Cand D of the formula C_(3)H_(6)Cl_(2) were isolated Each was Futherchorinated toprovetrichloro products (C_(3)H_(5)Cl_(3)) it wasfoundthat A provided one trichoro Product ,B gave two and C and D each gave three whatt are the structural formula of A, B ,C and D?

Answer»

SOLUTION :Dichoro isomers,
`(A) {:(""Cl),(""|),(" "CH_(3)-C-CH_(3)),(""|) ,(""Cl),(""darrCl_(2)),(""Cl),(""|),(CH_(2)Cl-C-CH_(3)),(""|),(""Cl),("(one)"):} (B) {:(""ClCH_(2)CH_(2)CH_(2)Cl),(""darrCl_(2)),(1.Cl_(2)CHCH_(2)CH_(2)Cl_(2)),(2.ClCH_(2)ClCHCH_(2)Cl):}`
`( C) {:(CH_(3)CHClCH_(2)Cl),(""darrCl_(2)),(1.CH_(2)ClCHClCH_(2)Cl),(2.CH_(3)CHClCHCl_(2)),(3. CH_(3)C Cl_(2)CH_(2)Cl):} (D) {:(CH_(3)CH_(2)XHXl_(2)),(""darr Cl_(2)),(1. CH_(3)CHClCHCl_(2)),(2.CH_(3)CH_(2)CCl_(3)):}`
Discussion
8.

In the table below, ul(identify) which transformations are alpha-decays and which are beta-decays. |{:(,alpha-decay,beta-decay),(.^(226)Ra rarr^(222)Rn,,),(.^(222)Rn rarr^(218)Po,,),(.^(218)Po rarr^(214)Pb,,),(.^(214)Pb rarr^(214)Bi,,),(.^(214)Bi rarr^(214)Po,,),(.^(214)Po rarr^(210)Pb,,),(.^(210)Pb rarr^(210)Bi,,),(.^(210)Bi rarr^(210)Po,,),(^(210)Po rarr^(206)Pb,,):}|

Answer»


ANSWER :`(##RES_PHY_CHM_V01_XI_C02_E01_402_A01##)`
Discussion
9.

In the systemLaCI_3 (s) +H_2O(g) +"Heat" to LaCIO(s) +2HCI(g), equilibrium is established . More water vapour is added to re-establish the equilibrium . The pressure of water vapour is doubled . The factor by which pressure of HCI is changed is

Answer»

2
`SQRT2`
`SQRT3`
`SQRT5`

Answer :B
Discussion
10.

In the system, A(g)+ 2B(g) hArr C(g) starting from 0.276M of A and 0.552M of B, the equilibrium is attained. If equilibrium concentration of B is found to be 0.12 M, then equilibrium constant for the equilibrium is :

Answer»

25
1.46
250
`4 xx 10^(-2)`

ANSWER :C
Discussion
11.

In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH_(4)Cl with Ca(OH)_(2). The by-product obtained in this process is

Answer»

`CaCl_(2)`
NaCl
NaOH
`NaHCO_(3)`

Solution :`2NH_(4)Cl+Ca(OH)_(2)to2NH_(3)+CaCl_(2)+2H_(2)O`. THEREFORE, `CaCl_(2)` is the by PRODUCT.
Discussion
12.

In the synthesis of NH_3 by Haber's process, catalyst is ____and promoter is______.

Answer»

SOLUTION :Finely divided ion (CATALYST) and molybdenum (PROMOTER).
Discussion
13.

In the synthesis of ammonia from nitrogen and hydrogen gases, if 6xx10^(-2) mole of hydrogen disappears in 10 minutes, the number of moles of ammonia formed in 0.3 minutes is

Answer»

`1.8xx10^(-2)`
`1.2xx10^(-3)`
`4XX10^(-2)`
`3.6xx10^(-2)`

SOLUTION :`N_(2) + 3H_(2) to 2 NH_(3)`
Rate `= -(d[N_(2)])/(dt) = (6xx 10^(-2)mol)/(10 min) = 6 XX 10^(-3) mol min^(-1)`
Rate of formation of `NH_(3)`
`i.e., "" (d[NH_(3)])/(dt) = (2)/(3)(-(d[H_(2)])/(dt))`
`= (2)/(3) xx 6 xx 10^(-3) mol min^(-1)`
`= 4 xx 10^(-3) mol min^(-1)`
Now, if `dt = 0* 3 min`,
`NH_(3)` formed `= (4 xx 10^(-3) mol min^(-1))(0.3 min)`
`= 1. 2 xx 10^(-3) mol`
Discussion
14.

In the synthesis of ammonia by Haber process , if 60 moles of ammonia is obtained in one hour , then the rate of disapperance of Nitrogen is

Answer»

30 mol/min
6 mol/min
`0.5` mol/min
60 mol/min

SOLUTION :`N_(2) + 3H_(2) = 2NH_(3)`
rate = `- (d[N_(2)])/(dt) = -(1)/(3) (d[H_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt)`
Rate of disappearance of `N_(2)`
`= (1)/(2)` of rate of formation of `NH_(3)`
`= (1)/(2) xx (60 "mole")/(1"hour")=(1)/(2) xx (60)/(60)` mole/minute
= 0.5 mole/minute .
Discussion
15.

In the study of inversion of sucrose in presence of acid, if r_(0),r_(1)" and "r_(oo) represent the polarimetric readings at times 0, t and oo respectively, then at the 50% inversion, which of the following relationship will hold good ?

Answer»

`r_(t)=r_(0)+r_(oo)`
`r_(t)=(1)/(2)(r_(0)+r_(oo))`
`r_(t)=r_(0)-r_(oo)`
`r_(t)=(1)/(2)(r_(0)-r_(oo))`

SOLUTION :Rotation in time `t = (r_(0) -r_(t)) PROP x`
Rotation in time oo = (r_(0) -r_(oo)prop a`
50% inversion means `x = a//2`
i.e. `r_(0) -r_(t) = (r_(0) -r_(oo))/(2) or 2 r_(0) - 2 r_(t) = r_(0) - r_(00)`
or `2 r_(t) = r_(0) + r_(oo)`
Discussion
16.

In the study of chlorination of propane four products (A,B,C and D) (structural isomerism) of the formula C_3H_6Cl_2were isolated. Each was further chlorinated to provide trichloro products (C_3H_5Cl_3)It was found that A provide one trichloro produce, B gave two and C and D each gave three. It is found that D is optically active. Correct formula of the compound D is

Answer»

`CH_3"CCl"_2CH_3`
`Cl CH_2 Cl CH CH_2 Cl`
`CH_3 CH_2 CH Cl_2`
`Cl CH_2 CH Cl CH_3`

SOLUTION :`ClCH_2CHCl CH_3` . It is OPTICAL ACTIVE
Discussion
17.

In the study of chlorination of propane four products (A,B,C and D) (structural isomerism) of the formula C_3H_6Cl_2were isolated. Each was further chlorinated to provide trichloro products (C_3H_5Cl_3)It was found that A provide one trichloro produce, B gave two and C and D each gave three. It is found that D is optically active. Correct formula of the product of chlorination of B is

Answer»

`Cl_2 CH CH_2 CH_2 CL`
`Cl CH_2 Cl CH CH_2 Cl`
both a and B

SOLUTION :`Cl - CH_2 - oversetunderset(|)(Cl)(C )H - CH_2 - Cl`
Discussion
18.

In the study of chlorination of propane four products (A,B,C and D) (structural isomerism) of the formula C_3H_6Cl_2were isolated. Each was further chlorinated to provide trichloro products (C_3H_5Cl_3)It was found that A provide one trichloro produce, B gave two and C and D each gave three. It is found that D is optically active.Formula of the compound A is

Answer»



`CH_3CH_2CHCl_2`
`CH_3-oversetunderset(|)(CL)(C )H - CH_2Cl`

SOLUTION :
Discussion
19.

In the structure given below, the sites S_(1) and S_(2) represent

Answer»

both octahedral voids
both TETRAHEDRAL voids
`S_(1)` - octahedral VOID `S_(2)` - tetrahedral void
`S_(1)` tetrahedral void`S_(2)`octahedral void

Answer :C
Discussion
20.

In the structure of diborane :

Answer»

the `B-H` BONDS are ionic
there are TWO three - CENTRED two ELECTRONS bonds
there is a `B-B` bond
there are two - centred three electron

Answer :B
Discussion
21.

In the storage battery during charging

Answer»

`pH` of the electrolyte INCREASES
`pH` decreases
`pH` remains unchanged
`pH` mayincrease or decrease depends on the extent of changing

Solution :CONCENTRATION of `H_(2)SO_(4)` decreases. HENCE `pH` increases.
Discussion
22.

In the statements regarding P_(4) molecule I) The oxidation state is zero II) The covalency is 4 III) The P-hat(P)bond angle 60^(@) The correct combination is

Answer»

Only III is correct
I & III are correct
All are correct
I and II are correct

ANSWER :2
Discussion
23.

In the statement of leukaemia ______ is used.

Answer»

WHITE PHOSPHORUS
Red phosphorus
Scarlet phosphorus
`P^32` isotope

Answer :D
Discussion
24.

In the standardization of Na_(2)S_(2)O_(3) using K_(2)Cr_(2)O_(7) by iodometry, the equivalent weight of K_(2)Cr_(2)O_(7) is ("molecular weight")/(X). What is the vlaue of X.

Answer»


Solution :`{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O),(""2I^(-)RARR I_(2)+2e^(-)xx3),(bar(Cr_(2)O_(7)^(2-)+14H^(+)+6I^(-)rarr 2Cr^(3+)+3I_(2)+7H_(2)O)):}`
`"So. eq. wt. "=("Molecular w.t.")/(6)`
Discussion
25.

In the standarization of Na_(2)S_(2)O_(3)using K_(2)Cr_(2)O_(7)by iodometry , the equivalent weight of K_(2)Cr_(2)O_(7)is

Answer»

MOLECULAR WEIGHT /2
Molecular weight /6
Molecular weight /3
Same as molecular weigh

Answer :B
Discussion
26.

In the standardization of Na_(2)S_(2)O_(3) using K_(2)Cr_(2)O_(7) by iodometry, the equivalent weight of K_(2)Cr_(2)O_(7) is

Answer»

(MOLECULAR weight)/2
(molecular weight)/6
(molecular weight)/3
Same as molecular weight

Solution :HYPO solution are USED as the reducing agent with POTASSIUM dicromate solution.
`K_(2)OVERSET(+6)(Cr_(2))O_(7)=("molecular weight")/(6)`
Discussion
27.

In the spinel structure, oxides, ions are cubical-closest packed whereas 1/8th of tetrahedral voids are occupied by A^(2+) cation and 1/2 of octahedral voids are occupied by B^(3+)cations. The general formula of the compound having spinel structure is:

Answer»

`A_2B_2O_4`
`AB_2 O_4`
`A_2B_4O_2`
`A_4B_2O_2`

Solution :No. of `A^(2+) = 1/8 XX 8= 1`
No. of `B^(3+) = 1/2 xx 4 = 2`
No. of `O^(2-) = 8 xx 1/8 + 6 xx 1/2 = 4`
`(AB_2O_4)`
Discussion
28.

In the species O_(2), O_(2) ^(- ) and O_(2) ^(2-), the correct decreasing order of bond strength is given as

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`O_(2) GT O_(2) ^(+) gt O_(2) ^(-) gt O_(2) ^(2-)`
`O_(2) ^(+) gt O_(2) gt O _(2) ^(-) gt O_(2) ^(2-)`
`O_(2) ^(2-) gt O _(2) ^(-) gt O_(2) ^(+) gt O _(2)`
`O _(2) ^(-) gt O_(2) ^(2-) gt O_(2) gt O _(2) ^(+)`

ANSWER :B
Discussion
29.

In the solvolysis of which alkyl halide, the tendency of rearrangement is minimum-

Answer»

cis-1-chloro-2-methylcyclohexane
2-chloro-2-phenylpentane
2-chloro-3-phenylpentane
2-chloro-4-phenylpentane

ANSWER :B
Discussion
30.

In the Solvay process, the reaction : 2NH_(4)Cl+Ca(OH)_(2)toCaCl_(2)+2NH_(3)+2H_(2)O takes place in :

Answer»

CARBONATION tower
saturation tank
ammonia RECOVERY tower
filtration tank

Answer :C
Discussion
31.

In the solid state PCl_5 exists as

Answer»

`[PCl_4]^(-)` and `[PCl_6]^(+)` IONS
covalent `PCl_5` molecules only
`[PCl_4]^(+)` and `[PCl_6]^(-)` ions
covalent `P_2Cl_(10)` molecules only

SOLUTION :In solid state, for `PCl_5`, following observation is observed.
`2PCl_(5)
Discussion
32.

In the solid state which of the following is the structure of analine ?

Answer»

WIDTH="30%">

`##CHY_CHE_ORG_XII_P2_PYQ_17_E01_059_O03.png" width="30%">
`##CHY_CHE_ORG_XII_P2_PYQ_17_E01_059_O04.png" width="30%">

ANSWER :B
Discussion
33.

In the solid state nitrogen penotoxide exists as

Answer»

`NO_2^(-)` and `NO_3^(+)`
`NO^(+)` and `NO_2^(-)`
`NO_3^(-)` and `NO_2^(+)`
`NO_2^(+)` and `NO^(-)`

ANSWER :C
Discussion
34.

In the silver plating of copper, K[Ag(CN)_(2)] is used instead of AgNO_(3). The reason is

Answer»

a thin layer of Ag is FORMED on Cu
more voltage is REQUIRED
`Ag^(+)` ions are completely removed from solution
less avalability of `Ag^(+)` ions, as Cu cannot displace Ag, from `[Ag(CN)_(2)]^-` ion.

Solution :In silver plating `K[Ag(CN)_(2)]` is USED which PROVIDES constant and required supply of `Ag^(+)` ions as `Ag(CN)_(2)^(-)` is very stable. But if `AgNO_(3)` is used concentration of `[Ag^(+)]` in solution will be very large in that case Ag will be DEPOSITED at faster rate without any unit ormity.
Discussion
35.

In the set of the given reactions, acetic acid yielded a product C. CH_(3)COOH overset(PCl_(5))to A underset(anh. AlCl_(3))overset(C_(6)H_(6))to B underset("ether")overset(C_(2)H_(5)MgBr)to C The product C would be

Answer»

`CH_(3)CH(OH)C_(2)H_(5)`
`CH_(3)COC_(6)H_(5)`
`CH_(3)CH(OH)C_(6)H_(5)`
`CH_(3)-overset(C_(2)H_(5))overset(|)C(OH)C_(6)H_(5)`

SOLUTION :`CH_(3)COOH overset(PCl_(5))to underset((A))(CH_(3)COCL) underset(ANH. AlCl_(3))overset(C_(6)H_(6))to underset((B))(C_(6)H_(5)COCH_(3)) underset("ether")overset(C_(2)H_(5)MgBr)to C_(6)H_(5)-underset(C_(2)H_(5))underset(|)overset(OMgBr)overset(|)C-CH_(3) overset(H^(+) // H_(2)O)to underset((C))(C_(6)H_(5)-underset(C_(2)H_(5))underset(|)overset(OH)overset(|)C-CH_(3))`
Discussion
36.

In the series Sc(Z= 21) to Zn ( Z= 30),the enthalpy of atomisation of zinc is the lowest , i.e., 126 kJ mol^(-1) . Why ?

Answer»

SOLUTION :In the series , Sc to Zn , all ELEMENTS have one or more unpaired electrons except zinc which has no unpaired electron as its outer electronic CONFIGURATION is `3d^(10)4s^(2)`. Hence, atomic intermetallic bonding ( METAL - metal bonding) is WEAKEST in zinc. Therefore, enthalpy of atomisation is lowest.
Discussion
37.

In the series Sc (Z=2) to Zn (Z=30), the enthalpy of atomisation of zinc is the lowest, i.e., 126kJ mol^(-1). Why?

Answer»

SOLUTION :In zinc, the d-electrons are not involved in a metallic bonding as d-orbitals are OCCUPIED completely, while in all other elements in first transition series, d-electrons participates in metallic bonding. So, Zn hs LOWEST enthalpy of ATOMISATION
Discussion
38.

In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^(-1). Why ?

Answer»

SOLUTION :In the series, Sc to Zn, all elements have ONE or more unpaired ELECTRONS except zinc which has no unpaired electron, its outer electronic configuration being `3d^(10)4s^(2)`. Lower the number of unpaired electrons, lower is the metal-metal bonding. HENCE, metal-metal bonding is weakest in zinc. Therefore, ENTHALPY of atomisation is lowest.
Discussion
39.

In the series Sc (Z = 21) to Zn (Z = 30), enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ. Mol^(-1). Why ?

Answer»

SOLUTION :In the above mentioned SERIES Sc (Z = 21) to Zn (Z = 30), all the METAL atoms have one or more unpaired electrons in their outermost shell. On the contry, ZINC `(3d^(10)4s^(2))` has no unpaired electron in the outershell, leading to very poor interatomic interactions. Therefore, enthalpy of atomisation of zinc is the lowest.
Discussion
40.

In the series reaction Aoverset(K_(2))(rarr)Boverset(K_(2))(rarr)Coverset(K_(2))(rarr)D, if K_(1)gtK_(2)gtK_(3) then the rate determing step of the reaction is

Answer»

`ARARRB`
`CrarrD`
`BRARRC`
Any STEP

ANSWER :B
Discussion
41.

In the series of reaction C_(6)H_(5)NH_(2)underset(0-5^(@)C)overset(NaNO_(2)//HCl)rarrXunderset(CH_(2)O)overset(HNO_(2))rarrY+N_(2)+HClX and Y are respectively.

Answer»

`C_(6)H_(5)-N=N-C_(6)H_(5),C_(6)H_(5)N_(2)^(o+)CL^(o+)`
`C_(6)H_(5)N_(2)^(o+)Cl^(o+),C_(6)H_(5)-N=N-C_(6)H_(5)`
`C_(6)H_(5)N_(2)^(o+)Cl^(o+),C_(6)H_(5)NO_(2)`
`C_(6)H_(5)NO_(2),C_(6)H_(6)`

SOLUTION :`C_(6)H_(5)NH_(2)UNDERSET(0^(@)C-5^(@)C)overset(NaNO_(2)//HCl)rarrH_(5)N_(2)^(+)Cl^(-)underset(H_(2)O)overset(HNO_(2))rarr`
`C_(6)H_(5)NO_(2)+N_(2)+HCl`
Discussion
42.

In the series of reaction C_(6)H_(5)NH_(2)overset(NaNO_(2)//HCl)underset(0-5^(@)C)rarr X overset(HNO_(2))underset(H_(2)O)rarr Y+N_(2)+HCl, X and Y are respectively

Answer»

`C_(6)H_(5)-N=N-C_(6)H_(5), C_(6)H_(5)N_(2)^(+)Cl^(-)`
`C_(6)H_(5)N_(2)^(+)Cl^(-),C_(6)H_(5)-N=N-C_(6)H_(5)`
`C_(6)H_(5)N_(2)^(+)Cl^(-),C_(6)H_(5)NO_(2)`
`C_(6)H_(5)NO_(2),C_(6)H_(6)`

Solution :`C_(6)H_(5)NH_(2)overset(NaNO_(2)//HCl)UNDERSET(0^(@)C-5^(@)C)rarr underset((X))(C_(6)H_(5)N_(2)^(+)Cl^(-)) overset(HNO_(2))underset(H_(2)O)rarr underset((Y))(C_(6)H_(5)NO_(2))+N_(2)+HCl`
Discussion
43.

In the series ethane,ethylene and acetyene, the C-H bond energy is :

Answer»

The same in all the THREE compounds
Greatest in ethane
Greatest in ethylene
Greatest in acetylene

Answer :D
Discussion
44.

In the series ethane,ethylene and acetylene, the C-H bond energy is

Answer»

GREATEST in ethylene
Greatest in acetylene
Greatest in ethane
The same in all the THREE COMPOUNDS

ANSWER :A
Discussion
45.

In the sequences of the following ._(92)Xe^(238) overset(-alpha)to Y overset(-beta)toZ overset(-beta)toL overset(-nx)to._(84)M^(218) the value of n will be

Answer»

5
4
3
6

Answer :B
Discussion
46.

In the sequence of the following reactions CH_(3)OH overset(HI)to CH_(3)I overset(KCN)to CH_(3)OH overset("reduction")to X overset(HNO_(2))to Y X and Y are respectively

Answer»

`CH_(3)CH_(2)NH_(2) and CH_(3)CH_(2)OH`
`CH_(3)CH_(2)NH_(2) and CH_(3)COOH`
`CH_(3)CH_(2)OH and CH_(3)CHO`
`CH_(3)OCH_(3) and CH_(3)CHO`

Answer :A
Discussion
47.

In the series : ""_(88)^(226)Raoverset(-alpha)to""_(86)^(222)Rnoverset(-alpha)to""_(84)^(218)RaAoverset(-alpha)to""_(82)^(214)RaBoverset(-Beta)to""_(83)^(214)RaC Ra belongs to which period?

Answer»

V
VI
IV
II

Answer :A
Discussion
48.

In the sequence of the following nuclear reaction ""_(98)^(238) X overset(-alpha)(to) Y overset(-beta)(to) Z overset(-beta) (to)Loverset(n alpha) (to) ""_(90)^(218) M What is the value of n

Answer»

3
4
5
6

Solution :`""_(98) X^(238) OVERSET(- (""_(2) He^(4)))(to) ""_(96) Y^(234) overset(- (""_(1) beta^(0)))(to) ""_(97) Z^(234) overset(- (""_(1) beta^(0)))(to) ""_(98) Z^(234)`
`""_(98) Z^(234) to ""_(90) Na^(218) + N(""_(2) alpha^(4))`
`234 = 218 + 4N implies n = ((234 - 218)/(4)) = 4 , 98 = 90 + 2N implies n = (8)/(2) = 4`
Discussion
49.

In the sequence of reactions Fe^(3+) (aq) underset(SCN^(-))overset("Excess of")rarr underset(("A"))("Blood red colour")underset(F^(-))overset("Excess of")rarr underset(("B"))("Colourless") Identify A and B. Write their IUPAC names. Explain the hybridisation involved in B and calculate spin only magnetic moment of B.

Answer»

Solution :`Fe^(3+)(AQ)+3SCN^(-)rarr underset("Trithiocyanato IRON (III)")underset(("A"))underset("Blood RED colour")([Fe(SCN)_(3)])underset(F^(-))overset("Excess of")rarr underset("Hexafluoridoferrate (III)")underset(("B"))underset("Colourless")([FeF_(6)]^(3-))`
Hybridisation in `[FeF_(6)]^(3-)=sp^(3)d^(2) "" (becauseF^(-) " is a weak ligand")`
In `[FeF_(6)]^(3-), Fe^(3+)` has the configuration `3d^(5)`, i.e., it has 5 unpaired electrons.
`THEREFORE` Magnetic moment `= sqrt(n(n+2))=sqrt(5(5+2))=sqrt(35)=5.92` BM.
Discussion
50.

In the sequence of reactions the final product (B) is

Answer»




ANSWER :C
Discussion