UP PGT Chemistry Mock Test - 9 - UPTET MCQ

125 mL of 0.05 M AgN0₃ = 125 x 0.05 millimoles 

A → A⁺ + e^- is favourable since, IE of A is low.
B + e^- → B^- is favourable since, EA of B is high.
Thus, A⁺B^- is favoured by low IE of A and high EA of B.

(a) Zn is oxidised to Zn²⁺ in complex

(b) Au⁺ is reduced to Au — True
(c) Au (s) is displaced
(d) Since, it is a spontaneous reaction ΔG° < 0

• An amorphous or non-crystalline solid is a solid that lacks the long-range order that is characteristic of a crystal.
• In brief, an amorphous solid is any noncrystalline solid in which the atoms and molecules are not organized in a definite lattice pattern. Such solids include glass, plastic, and gel.
• Quartz glass (SiO₂) is an amorphous solid due to its short range order of constituent particles. Note Quartz is a crystalline solid while quartz glass is an amorphous solid.

It is an α, β-unsaturated ketone which can be formed in an aldol condensation followed by dehydration.

• An ionic compound is a giant structure of ions.
• The ions have a regular, repeating arrangement called an ionic lattice .
• The lattice is formed because the ions attract each other and form a regular pattern with oppositely charged ions next to each other.

Plutonium (Z = 94) have [Rn] 5f⁶6d⁰7s² configuration and show oxidation state upto +7.

Like dissolves like. Polar solvents (as H₂O) dissolve polar compounds (ionic compounds) and non-polar solvents (as ether) dissolve non-polar compounds (covalent compounds).
By Fajans’ rule smaller the size of cation, larger the size of anion. Larger the charge on cation or anion then larger the polarising power and thus, larger the covalent nature and thus, solubility in non-polar solvents.

BCl₃ + 3H ⇋ B+3HCl
BCl₃ + 3/2 → B + 3HCl
3/2 mol → 1 mol
1 mol B → 3/2 molH₂
2 mole → 3/2  x 2 = 6/2 = 3 mole
1 mole = 22.4 liter S.T.P
3----? 
= 22.4 x 3
=67.2 litre

Decomposition of N₂O₅ is 1st order reaction.

Salt of weak acid & weak base

(a,c)

(a) Vapour pressure of a component in a solution,  by Raoult's law, thus correct.

(b) π = i MRT

For NaBr,

y = 2, / = 1 + (y - 1)x = 2

M = 1, T = (273 + t )

∴  π = 2R (273 + t) thus, incorrect.

(c)

π = i MRT

π  ∝ for equimolar solution

i = 1 for sucrose

i = (1 + x)forCH₃COOH (x < < 1)

for CH₃COOH being weak acid

i = 2 for KCI

i = 3 for K₂S0₄

Thus, π (osmotic pressure) of 0.01 M solutions of sucrose < CH₃COOH < KCI < K₂S0₄ Thus, correct.

 and K₁ are different, hence ΔT_f will be different, thus incorrect.

Also X, has CH₃—CH(OH)—, gives iodoform test and it is chiral.

Diazonium reacts with phenols and napthols to form dyes.

When freshly prepared solution of FeSO₄ is added in a solution containing NO₃^– ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.

Periods:

• Elements are arranged in increasing the atomic number of elements in a period.
• One extra electron gets added to the outermost shell as we move along the periods from left to right.
• The electron gets added to the same shell or orbit and thus the electrons present for bonding increase by one unit.
• Thus, the shell number remains the same but the number of electrons present for bonding increases along a period.

Groups:

• Elements having the same number of outer electrons are put in the same group of the periodic table.
• When we move down a group, one extra shell gets added to the elements.
• The outermost shell has electrons present for bonding.
• Though the number of shells increases as we go down in a group, the number of electrons in the outermost shell remains the same.
• For example, the Halogens F, Cl, Br, I, At all belong to group 17 and have 7 electrons in the outermost shell.
• Similarly, Group 16 elements have 6 electrons in the outermost shell, group 15 has 5 electrons in the outermost shell, and so on.

Hence, group 17 is called halogens.

(b)

Structure is cubic close packing.
Number of atoms of M = (from edge centre) +1 (from body centre) =2

Number of atoms of X =   (from corners of the cube) +  (from face centre)  =4

Thus, M : X = 2 : 4 = 1 : 2 Hence, the empirical formula of the compound is MX₂.

Required reaction= Heat of formation of water – H2(g) +1/2O₂ (g)= H₂O(l)
WRITE THE EQUATION PROPERLY FOR EACH EQUATION
 rearranging the reactions we have to get the above equation
B.E. of H--H= H₂(g) → 2H(g)..........X1
B.E of O₂ = O₂(g) ---> 2O(g)...........X2
B.E. OF O-H = O-H(aq)--> O(g)+H(g) …......X3
LATENT HEAT OF VAPOURISATION OF WATER = H₂O(l)----> H₂O(g)..........X4
THUS REVERSING EQTN X₃ AND X₄ , HALVING EQUATION X₂ AND DOUBLING EQUATION X₃ WE WILL GET THE DESIRED EQUATION.

Both are bent shape molecule.

(a) 

Oxidation takes place at anode - correct

2H₂O(l) →4H⁺ (aq) + O₂(g) + 4e^-

(b) Reduction take place at cathode - correct

since, E⁰_cell < 0, in electrolysis.

Cell reaction is spontaneous.

(c) Copper is deposited at the cathode, hence the meted to be electroplated (say Fe), should be made cathode - correct (d) Since. Cu²⁺(aq)- (blue) is reduced Cu(s), hence blue colour fades - correct. 
 

The correct answer is option D
Alkyl halides react with dialkyl copper reagents to give alkanes. The reaction is called Corey House synthesis.
R2​CuLi+R′X→R−R′+RCu + LiX
For example, ethyl iodide reacts with lithium diethyl cuprate to form butane.
(C_2​H₅​)₂​CuLi+C₂​H₅​−I→C₂​H₅​−C₂​H₅​+C₂​H₅​Cu + LiI

AB₃E: trigonal pyramidal (central atom + 3 outer atoms make a pyramid)
→ start with AB₄ molecule (tetrahedral) and replace a B atom w/ lone pair
→ lone pair electrons push bonding electrons away
→ bond angles are now less than 109.5°

dsp² is square planar geometry.

The correct answer is Option A.

NH₄OH + H+ ⇌ NH₄+ + H_2​O
∴
Again,
NH₃ + H₂O → NH₄OH ⇌ NH
4++ OH-

 = [OH
-] [H+] (∵H2​O is in excess)
       =K_w
          =1×10^-14
∴K = K×1×10^-14
        =1.8×10⁹×10^-14
        =1.8×10^-5

Water has a hybridization of sp³. In a water molecule oxygen has two lone pairs and two bonded hydrogens. Its structure is V in shape and has a bent structure. Water has a chemical formula of H₂O.

(8) Barium phosphate is Ba₃( P0₄)₂.