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.

Specific heat of a metal is 0.115 cal K^(-1) . If the weight percentage of metal in metal oxide is 77.28, determine the valency of metal.

Answer»

Solution :Approximate ATOMIC mass of metal `=(6.4)/("specific HEAT")=(6.4)/(0.115)=55.65`
In 100 G of metal oxide, weight of metal =77.28g
Weight of oxygen =22.22g
`(W_(i))/(W_(2))=(E_(1))/(E_(2))`
Equivalent weight of oxygen =8
Equivalent weight of metal `=(77.28)/(22.22)xx8=27.82`
Valency of meta `=("approximate atomic weight ")/("equivalent weight") =(55.62)/(27.82)=2`
2.

Specific acticvity (activiy per gram) of a sample of .^(239)Pu and .^(240)Pu was found to be 6xx10^(9) dps. Given that t_(1//2) (Pu - 239) and t_(1//2) (Pu - 240) are 2.44xx10^(4) year and 6.58xx10^(3) year respectively, then calcualte the isotopic compostion of mixture.

Answer»

Solution :Specific activity expresses that `1g` sample shows the activity `6xx10^(9)` dps. Let the mixture CONTAIN `a g Pu-239` and `b g Pu-240`, then
`a+b = 1` ....(1)
Now, separately for `Pu-239` and `Pu-240`, the RATE of DECAY is `r_(1)` and `r_(2)`respectively.
Then, `r_(1) = lambda.N = (0.693)/(t_(1//2)) xx (a xx N_(A))/(239) dpsg^(-1)`
`= (0.693xx a xx 6.023xx10^(23))/(2.44xx10^(4)xx 365xx24xx60xx60xx239)`
`= 2.27xx10^(9)xx a dpsg^(-1)`
SIMILARLY, `r_(2) = (0.693)/(t_(1//2)) xx (b)/(240) xx N_(A) dpsg^(-1)`
`= 8.38xx10^(9)xx b dpsg^(-1)`
But, `r = r_(1) + r_(2)`
`2.27xx10^(9) a + 8.38 xx 10^(9)b = 6xx10^(9)`
or `2.27a + 8.38b = 6` ...(2)
Solving equations (1) and (2),
`a = 0.39` or `39%`
`b = 0.61` or `61%`
3.

Species having same bond order are :

Answer»

`N_(2)`
`N_(2)^(-)`
`F_(2)^(+)`
`O_(2)^(-)`

Solution :(C, D)
Bond order of the following SPECIES are calculated using MOLECULAR orbital electronic CONFIGURATION and OBTAIN as`N_(2) = 3 , N_(2)^(-) = 2.5 , F_(2)^(+) = 1.5 , O_(2)^(-)`= 1.5
4.

Sparingly soluble salts maintains their solubilityproduct value in their saturated solutions irrespective of the sources of the ions .what is the molar solubility of AgNO_3in a 0.1 M H_2Ssolution buffered at pH= 2 (K_1 and K_2 "of "H_2S " are "10^(4) and 10^(-8)respectively)(K_(sp) " of "Ag_2S = 4xx 10^(-13) )(Note : No Ag_2Sprecipitate should be formed)

Answer»

` 0.01 M`
` 0. 02 M`
` 0. 03 M`
` 0. 04 M`

Solution :`PH= 2 , [H^(+) ] =10 ^(-2) M`
` K_a =([H^(+) ][S^(-2) ])/( [H_2S] ) rArr 10 ^(-12)=( 10 ^(-4) [S^(-2)])/( 10^(-1) ) `
` [S^(-2) ]= 10 ^(-9)M`
` Ag_2S hArr 2Ag^(+) +S^(_2) `
` K_(Sp)=[Ag^(+) ]^(2) [S^(-2) ], 4 xx10 ^(-13) =[Ag^(+) ]^(2) (10 ^(-9))`
` [Ag^(+) ]= 2xx 10 ^(-2) =0.02 M`
5.

Sparingly soluble salts maintains their solubilityproduct value in their saturated solutions irrespective of the sources of the ions .What is the solubilityof MgCl_2 (gm// L)in a 0.01 M KOHsolution without causing precipitation of Mg(OH)_2 ?K_(sp) "of "Mg (OH)_2 = 2xx 10^(8)

Answer»

`1.9 xx 10^(-2) `
` 2.8 xx 10^(-3) `
` 3.4 xx 10^(-1) `
` 4.6 xx 10^(-4) `

SOLUTION :` [KOH ]= 0.01 M , [OH^(-) ] =0.01 M `
` K_(sp) =[Mg^(+2) ] [OH^(-) ]^(2) , 2 xx10 ^(-8)=[Mg^(+2) ](10 ^(-2) )^(2) `
` [Mg^(+2) ] =2 xx 10 ^(-4)" mol"//"lt" `
` [MGCL _2] =2 xx 10 ^(-4) M= 2 xx 10 ^(-4)xx 95 GM //LIT `
` ""= 1.9 xx 10 ^(-2)gm//lit `
6.

sp^(2) Hybrid orbitals are not present in

Answer»

`so_2`
`BF_3`
`B_2 H_6`
`SO_3`

ANSWER :C
7.

What is Sorel's cement and how is it prepared?

Answer»

`5MgCl_2 . MGO . xH_2O`
`MgCl_2 . 5 MgO . xH_2O`
`MgCl_2 . MgO . xH_2O`
`MgCl_2 . 2MgO . xH_2O`

SOLUTION :`MgCl_2 . 5 MgO . xH_2O`
8.

Soperoctet molecule is

Answer»

`CiF_(3)`
`NH_(3)`
`PCl_(3)`
`CO_(2)`

SOLUTION :`10 e^(-)`are PRESENT around Cl.
9.

Sometimes a red colour is not produced in the Lassaigne's test even if both nitrogen and sulphur are present in the organic compound. Explain.

Answer»

Solution :In principle, if the organic compound contains both N and S, sodium thiocyanate should be formed in Lassaigne's test and this should give BLOOD red colouration with `FeCl_(3)`
`{:(Na+,underset("(From organic compound)")(C+N+S),overset("Fusion")rarr,underset("Sod. thiocyanate")(NACNS),),(,3NaCNS+FeCl_(3),rarr,underset(underset("(Blood red colouration)")("FERRIC thiocyanate"))(Fe(SCN)_(3)),+ 3NaCl):}`
However, if blood red colouration is not obtained, it does not necessariy mean that S is absent. This is because in presence of excess of sodium METAL, sodium thiocynanate initially formed, decomposes to FORM sodium cyanide and sodium sulphide.
`2Na + NaCNS overset(Delta)rarr NaCN + Na_(2)S`
As a result, blood red colouration is not obtained.
10.

Some times a reacant undergoes chemical/radioactive changes followingtwo or more different pathsto yield twoor more different producesrespectively. Such reactions are called parallel path reactions. If K_(1) and K_(2) are rate constans for the reaction of A follwing two parallel paths, then Then K_(av) = K_(1) + K_(2) Which one of the following is correct for the above reaction:

Answer»

Fractionl YIELD of `p = (K_(1))/(K_(2))`
FRACTIONAL yield of `p = (K_(1) + K_(2))/(K_(av))`
Fractional yeild of `p = (K_(1))/(K_(av))`
Fractional yield of `p = (K_(2))/(K_(av))`

Solution :Use `K_(av) = K_(1) + K_(2)`
11.

Some times a reacant undergoes chemical/radioactive changes followingtwo or more different pathsto yield twoor more different producesrespectively. Such reactions are called parallel path reactions. If K_(1) and K_(2) are rate constans for the reaction of A follwing two parallel paths, then Then K_(av) = K_(1) + K_(2) For if E_(1) and E_(2) are energy fo activations, then

Answer»

`E_("Total") = E_(1) + E_(2)`
`E_("Total") = E_(1) - E_(2)`
`E_("Total") = K_(1) E_(1) + K_(2)E_(2)`
`E_("Total") = (K_(1) E_(1) + K_(2)E_(2))/(K_(1) + K_(2))`

Solution :`K_(av) = K_(1) + K_(2)`
`A_(T)E^(-E_(T)//RT^(2)) = A_(1) e^(-E_(1)//RT^(2)) + A_(2) e^(-E_(2)//RT^(2))` ....(1)
DIFFERENTIATING EQN. (1)
`A_(T). (E_(T))/(RT).e^(-E_(T)//RT^(2)) = A_(1). (E_(1))/(RT).e^(-E_(1)//RT^(2)) + A_(2) (E_(2))/(RT) e^(-E_(2)//RT^(2))`
`K_(av) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)`
`(K_(1) + K_(2)) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)`
`:. E_(T) = (K_(1) E_(1) + K_(2) E_(2))/(K_(1) + K_(2))`
12.

Some times a reacant undergoes chemical/radioactive changes followingtwo or more different pathsto yield twoor more different producesrespectively. Such reactions are called parallel path reactions. If K_(1) and K_(2) are rate constans for the reaction of A follwing two parallel paths, then Then K_(av) = K_(1) + K_(2) If average life of A for p is T_(1) and for Q is T_(2) then:

Answer»

`T_(AV) = T_(1) + T_(2)`
`T_(av) = (T_(1)T_(2))/(T_(1) + T_(2))`
`T_(av) = (T_(1) + T_(2))/(T_(1)T_(2))`
`T_(av) = K_(1) T_(1) + K_(2) T_(2)`

SOLUTION :`K_(av) = (1)/(T_(av))`
13.

Some subtances posses some residual entropy even at absolute zero. This is because

Answer»

they contain some impurities
they attain different orientationsof moleculeseven at ABSOLUTE zero
it is DIFFICULT to attain absolute zero of temperature
all the above FACTORS are responsible

Solution :The residual ENTROPY at absolute zero is due to different ORIENTATION of the molecules even at absolute zero.
14.

Some time ago formation of polar stratospheric clouds was reported over Antarctica. Why were these formed ? What happens when such clouds break up by warmth of sunlight?

Answer»

Solution :In summer season, `NO_2` and `CH_4` react with chlorine monoxide and chlorine atoms forming chlorine sinks, preventing much ozone depletion.
Whereas in winter, special type of CLOUDS called polar stratospheric clouds are formed over Antarctica. These polar stratospheric clouds provide surface on which chlorine nitrate gets hydrolysed to form hypochlorous acid. It ALSO reacts with HYDROGEN chloride to give molecular chlorine.
`ClO_((g))^* + NO_(2(g)) to underset("Chlorine nitrate")(ClO NO_(2(g))`
`Cl_((g))^* + CH_(4(g)) to ""^*CH_(3(g)) + HCl_((g))`
`ClONO_(2(g)) + H_2O_((g)) overset("Hydrolysis")to HOCl_((g)) + HNO_(3(g))`
When SUNLIGHT returns to the antarctica in the spring, the sun.s warmth breaks up the clouds and `HOCI, Cl_2` are photolysed by sunlight.
`HOCl_((g)) overset(hv)to O^* H_((g)) + Cl_((g))^*`
`Cl_(2(g)) overset(hv)to 2Cl_((g))^*`
The chlorine radicals thus formed, INITIATE the chain reaction for ozone depletion.
15.

Some substances behave as electrolytes in dilute solutions and as colloids in their concentrated solutions. Their colloidal forms are sait to form

Answer»

Emulsions
Gels
Micelles
Sols

Answer :C
16.

Some statements regarding air pollution are given. Among them, the correct statements are (a) Above 80% of CO is released from automobiles. (b)in urban areas, at the peak time of the traffic the level of CO is 100 to 500ppm (c) If the percentage of CO-Hb in the blood is 32% it causes immediate death (d) TLV of CO in the atmosphere is 9ppm

Answer»

All ONE correct
Only a,B and d
Only a and d
Only b, C and d

Answer :B
17.

Some statements are given with regard to entropy. The incorrect statement(s) are (A) The absolute entropy of substances cannot be determined (B) In standard state entropy of elements is always positive (C ) The entropy of universe always decreases (D) In a spontaneous process, for an isolated system the entropy of the system generally increases

Answer»

A, B
B,C
A,C
only C

Solution :Absolute Entropy of SYSTEM can be CALCULATED from THIRD LAW of Thermodynamics. As per SECOND law of Thermodynamics Entropy of universetends to increases
18.

Some statements are given regarding nature of oxides (i) In second period, nitrogen form strongest acidic oxide (ii) In third period, sodium forms strongest basic oxide (iii)Oxides of metalloids are generally amphoteric in nature

Answer»

I and II are CORRECT
II and III are correct
I and III are correct
I, II and III are correct

ANSWER :A
19.

Some statements are given regarding IVA group elements A) Order of Electronegativity : C> Si = Ge=Sn Si > Ge > Pb > Sn C) Order of Melting point : C> Si > Ge > Pb > Sn Correct orders of the above

Answer»

A only
A,B only
B,C only
A,B,C

Answer :D
20.

Some statements are given below. Among them, the correct statement are a) All transition elements are d - block elements b) All d- block elements are transition elements c) Helium is a inert gas d) Aluminium is a respresentative element

Answer»

All are CORRECT
Only a, C and d are correct
Only B,c and d are correct
Only a and c are correct

Answer :B
21.

Some statements are give below: (i) The formation of a cation from a neutral atom is favoured by small size of the atom (ii) pi bond does not exist between two atoms withot sigma bond (iii) The formation of chemical bond is associated with an increase in potentialenergy.

Answer»

only I and II are correct
only ii is correct
only ii and III are correct
only I and iii are correct

Answer :B
22.

Some statements are given. Among them the correct statements are (a) IP_2 of sodium is greater than that of Magnesium (b) IP_2 of lithium is greater than IP_1 of Helium (c) IP_2 of sodium is greater than IP_1 of Neon (d) IP_1 of oxygen is greater than that of Nitrogen

Answer»

All are CORRECT 
Only a, B and C are correct 
Only a and b are correct
Only a and d are correct 

ANSWER :B
23.

Some statements about valence bond theory are given below (i) The strength of bond depends upon extent of overlapping. (ii) The theory explains the directional nature of covalent bond. (iii) According to this theory oxygen molecule is paramagnetic in nature.

Answer»

all are correct
only i and III are correct
only i and II are correct
all are WRONG

Answer :C
24.

Some statements about the structur of diborane are given below . NMR and RAMANspectral studies have confirmed that four hydroens of diborance are one tuype and remaing two are of another type B) Electron diffraction studies have shown that diborance contains two copanar BH_(2) groupsc) Diborane is a planar molecule D) Boronof dibrane ujndergoes sp^(2) hybridisation . the correct statement are

Answer»

Only A and B
Only A,B,C
Only B,C,D
All are correct

Answer :1
25.

Some statements about the structure of diborane are given below(A) Studies have confirmed that four hydrogens of diborane are one type and remaining two are of another type (B)Diborane contains two coplanar BH_(2)groups (C)Diborane is a planar molecule(D)Boron of diborane undergoes sp^(2)hybridization The correct statements above are

Answer»

Only A, B, C
Only A and B
Only B, C, D
All are correct

Answer :A
26.

Some statements about heavy water are given below. (i) Heavy water is used as a moderator in nuclear reactors (ii) Heavy water is more associated than ordinary water. (iii) Heavy water is more effective solvent than ordinary water Which of the above statements are correct ?

Answer»

(i) and (II)
(i), (ii) and (III)
(ii) and (iii)
(ii) and (iii)

ANSWER :A
27.

some statement regarding the reaction: are given below. Select the correct statement(s).

Answer»

P is a TRANS alkene
`Q_(1)` is a PURE compound and optically inactive due to internal
In the P to `Q_(1)` conversion step and `Br_(2)` adds on P in a sy manner and the intermediate FORMED is a cyclic brominium ION.
`Q_2` is a BINARY mixture and is optically inactive.

Answer :A::D
28.

Some solid NH_(4) HS " is placed in a flask containing " NH_(3) " at " 0*5 " atm. What would be the pressure of " NH_(3) and H_(2)S " when equilibrium is reached " ? NH_(4) HS (s) hArr NH_(3) (g) + H_(2) S (g), K_(p) = 0*11

Answer»

<P>

Solution :`NH_(3) and H_(2)S ` PRODUCED by the decomposition of `NH_(4)HS`will be same . Suppose at equilibrium each has pressure= p atm due to decompistion of ` NH_(4)HS` . Then
` p_(H_(2)S) = p atm `
` p_(NH_(3) ) p + 0*5" "(:' NH_(3) " is already presentat" 0*5atm )`
Applying law of chemical equilibrium to the given reaction
`p_(NH_(3)) xx p_(H_(2)S) = K_(p)`
`:.p xx(p + 0*5)= 0*11`
or ` p^(2) + 0*5 p = 0*11`
or`p^(2) + 0*5 p - 0*11 = 0 `
` :. p= -0*5pm sqrt((0*5)^(2)-4 (-0*11))/2 = (0*5 pm sqrt (0*25 + 0*44))/2= -(0*5 pm sqrt(0*69))/2`
` ( -0*5 pm 0*83 )/2 = (0*33)/2 = 0*165 ("Neglecting-ve value ")`
` :. p_(H_(2)S) = 0* 165 "atm " , p_(NH_(3))= 0*665 " atm "`
29.

Some solid NH_4HS is placed in a flask containing 0.5 atm NH_3. What is the equilibrium presence of NH_3 if K_p =0.11 atm^2 for the reaction NH_4HS(s) leftrightarrow NH_3(s)+H_2S(g).

Answer»

SOLUTION :0.665 ATM
30.

Some reasons are given regarding the limited use of H_(2) as fuel I) Its calorific value is low II) Its availability in free state is less III) Its transportation is easy The correct statements are

Answer»

I, II and III
II, III and IV
All are CORRECT
II and III

Answer :D
31.

Some reactions yield greater amount of products on heating while some others give lesser amount. Why ?

Answer»

Solution :On increasing the temperature, endothermic reactions SHIFT FORWARD while exothermic reactions shift BACKWARD ( by LE Chatelier's principle ).
32.

Some process are given below. What happens to the process if it is subjected to a change given in the barckets ? (ii) Dissolution ofIceoverset(M.pt) hArr " Water ( Pressure is increased ) " (ii) Dissolution of NaOH in water ( Temperature is increased) (iii)N_(2) (g) + O_(2) (g) hArr 2 NO (g) - 180* 7 kJ(pressure is increased and temperature is decreased).

Answer»

SOLUTION :(i)
Equilibrium will SHIFT in the forward direction , i.e., more of ICE will melt.
(ii) Solubility will decrease because it is an exothermic process.
(iii) Pressurehas no effect. Decrease of TEMPERATURE will shift the equilibrium in the backward direction.
33.

Some oxidation reactions of emthane are given below. Which of them is/are controlled oxidation reactions ?

Answer»

`CH_(4(g)) + 2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((L))`
`CH_(4(g)) + O_(2(g)) rarr C_((s)) + 2H_(2)O_((l))`
`CH_(4(g)) + O_(2(g)) overset(Mo_(2)O_(3))rarr HCHO + H_(2)O`
`2CH_(4(g)) + O_(2(g)) overset("Cu/523/100 atm")rarr 2CH_(3)OH`

Solution :Alkanes on COMBUSTION give `CO_(2)` and `H_(2)O`. Combustion in insufficient supply of air or `O_(2)` given CARBON black and water. Alkanes on heating with regular supply of oxygen or in controlled WAY give HCHO and `CH_(3)OH`.
`CH_(4(g))+O_(2(g))overset(Mo_(2)O_(3))rarr HCHO + H_(2)O`
`2HC_(4) + O_(2(g)) overset("Cu, 523 K, 100 atm")rarr 2CH_(3)OH`
34.

Some oxidation reactions of methane are given below. Which of them is/are controlled oxidation reactions ?

Answer»

`CH_4 (g) +2O_2(g) to CO_2 (g) + 2H_2O (l)`
`CH_4(g) + O_2 (g) to C (s) + 2H_2O(l)`
`CH_4(g) + O_2(g) overset(Mo_2O_3)to HCHO + H_2O`
`2CH_4(g) + O_2 overset"Cu/523 K/100 atm"to 2CH_3OH`

SOLUTION :Reaction (c ) and (d) in which `CH_4` does not UNDERGO complete combustion to give `CO_2` and `H_2O` or INCOMPLETE combustion to C and `H_2O` are CONTROLLED oxidation reactions.
35.

Some oxidation reactions of methane are given below. Which of them is/ are controlled oxidation reactions?

Answer»

`CH_(4)+ 2O_(2) to CO_(2)(g)+ 2H_(2)O(l)`
`CH_(4)+ O_(2) to C(s)+ 2H_(2)O(l)`
`CH_(4)+ O_(2) OVERSET(Mo_(2)O_(3)) to HCHO+ H_(2)O`
`2CH_(4)+O_(2) overset("Cu//523K//100atm")to 2CH_(3)OH`

Solution :In which `CH_(4)`, does not undergo complete combustion to give `CO_(2)` and `H_(2)O`.
36.

Some of the properties of water are described below. Which of them is/are not correct /

Answer»

Water is known to be a UNIVERSAL solvent
Hydrogen bonding is present to a LARGE extent in liquid water.
There is no hydrogen bonding in the frozen STATE of water
Frozen water is heavier than liquid water.

Solution :There is H-bonding EVEN in frozen water , and frozen water , i.e., ICE is lighter than liquid water.
37.

Some organic compounds are purified by distillation at low pressure because the compounds are

Answer»

are LOW BOILING liquids
are high boiling liquids
are highly volatile
decompose at their NORMAL boiling point

ANSWER :D
38.

Some of the properties of the two species, NO_(3)^(-) and H_(3)O^(*) are described below. Which one of them is correct ?

Answer»

SIMILAR in HYBRIDIZATION for the CENTRAL atom with different structures.
Dissimilar in hybridization for the central atom with different structures.
Isostructural with same hybridization for the central atom.
lsostructural with different hybridization for the central atom.

SOLUTION :Dissimilar in hybridization for the central atom with different structures.
IN `NO_(3)^(-2) `, nitrogen is in `sp^(2)` hybridization, thus planar in shape. In `H_(3)O^(+)`, oxygen is in `sp^(3)` hybridization , thus tetrahedral in shape.
`therefore ` Correct answer : (B)
39.

Some of the properties of the two species ,NO_(3)^(-) and H_(3)O^(+)are decribed below . Which is notof them is correct ?

Answer»

Dissimilar in hybridization for the central
atom with DIFFERENT strcutures
Isostructural with same hybridization for the
centralatom
Isostructural with different hybridization for
the central atom
Similar in hybridization for the central atom
with different structures

Solution :Applying the formula,
`X= (1)/(2) [VE + MA - c + a]`
For `NO_(3)^(-) , X = (1)/(2) [5 + 0 - 0 + 1] = 3 `
HYBRIDISATION = ` sp^(2)` , Structure = Planar triangular
For `H_(3)O^(+) , X = (1)/(2) [ 6 + 3 - 1 + 0 ] = 4 `
Hybridisation = `sp^(3)`, Structure= PYRAMIDAL
Thus, hybridisation as well as structures are different.
40.

Some of the properties of water are described below. Which of them is/are not correct ?

Answer»

Water is known to be a universal solvent
Hydrogen bonding is present to a large extent in liquid water.
There is no hydrogen bonding in the frozen STATE of water.
Frozen water is heavier than liquid water.

Solution :There is H-bonding in frozen water so, ice is lighter than liquid water.
The crystalline form of water is ice. At ATMOSPHERIC PRESSURE, ice CRYSTALLISES in the hexagonal form, but at very low temperatures it condenses to cubic form. Density of ice is less than that of water. Therefore, an ice cube FLOATS on water.
41.

Some of the major used of heavy water are given below. Which one is not correct?

Answer»

It is USED as a moderator in nuclear reactors.
It is used as a TRACER compound for studying reaction mechanism.
HIGH concentration of HEAVY water accelerates the growth of plants.
It is used in preparing deuterium.

Solution :High concentration of heavy water retards the growth of the plants.
42.

Some of the Group 2 metal halides are covalent and soluble in organic solvents . Among the following metal halides , the one which is soluble in ethanol is

Answer»

`BeCl_(2)`
`MgCl_(2)`
`CaCl_(2)`
`SrCl_(2)`

Solution :Due to small size , high IONIZATION enthalpy of Be , `BeCl_(2)` is covalent and hence most soluble in ORGANIC SOLVENTS such as ethanol .
43.

Some of the Group-2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is…

Answer»

`BeCl_(2)`
`MgCl_(2)`
`CaCl_(2)`
`SrCl_(2)`

SOLUTION :`BeCl_(2)`has covalent CHARACTER because of small size and high EFFECTIVE nuclear CHARGE. Being covalent in nature it is dissolved in ETHANOL.
44.

Some of the Group 2 metal halides are covalent and soluble in organic solvents. Among the following metal halides, the one which is soluble in ethanol is _____________

Answer»

`BeCl_2`
`MgCl_2`
`CaCl_2`
`SrCl_2`

ANSWER :A
45.

Some of the following properties of two species, NO_(3)^(-)" and "H_(3)O^(+) are described below. Which one of them is correct ?

Answer»

dissimilar in HYBRIDISATION for the central atom with DIFFERENT structure.
isostructural with same hybridisation for the Central atom.
different hybridisation for the central atom with same structure
none of these

Solution :`NO_(3)^(-) - SP^(2)` hybridisation, planar
`H_(3)O^(+)-sp^(3)` hybridisation, pyramidal
46.

Some meta-directing substituents in aromatic substitution are given,Which one is most deactivating?

Answer»

`-COOH`
`-NO_(2)`
`-C-=N`
`-SO_(3)H`

ANSWER :b
47.

Some meta-directing substituents in aromatic substitutionare given.Which one is most deactivating ?

Answer»

`-COOH`
`-NO_2`
`-C-=N`
`-SO_3H`

ANSWER :B
48.

Some meta - directing substituents in aromatic substitution are given. Which one is most deactivating?

Answer»

`- SO_3H`
`- COOH`
`- NO_2`
`-C -= N`

ANSWER :C
49.

Some meta directing substituents in aromatic substitution are given which one is most deactivating?

Answer»

`-COOH`
`-NO_2`
`-C-=N`
`-SO_3H`

SOLUTION :`-NO_2`
50.

Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating

Answer»

`-COOH`
`-NO_(2)`
`-C -=N`
`-SO_(3)H`

ANSWER :B