Find the final product

Question: What will be the final product P of the following reaction?

Solution: To find out the final product let us treat the given reactant with the reagent step by step:

First the given reactant is treated with H₃O⁺.

[H₃0]+ is a hydronium cation, also known as hydroxonium. It's an acid according to Bronstend-Lowry theory, about bases and acids, because it can give of a [H]⁺ :

[H₃0] <--> H₂O + H⁺

So when given reactant reacts with H₃O⁺, it gives:

On heating the obtained product losses CO₂

Now the obtained alkene is react with O₃ and H₂O₂.

As a reagent, ozone is used to cleave alkenes. When alkenes are treated with ozone and H₂O₂, they are subjected to “Oxidative workup”, this breaks the carbon-carbon double bond to form carboxylic acid. 

Now this is final product P. 

Find the standard enthalpy of combustion

Question: The standard enthalpies of formation of CO₂ (g), H₂O (l) and glucose (s) at 25^oC are -400 kJ/mol, -300kJ/mol and -1300kJ/mol, respectively. What will be the standard enthalpy of combustion per gram glucose at 25^oC?

Solution: First let us know the meaning of standard enthalpy of combustion.

The standard enthalpy change of combustion is the enthalpy change which occurs when one mole of an element or compound is burned completely in oxygen under standard conditions.

These standard conditions are:

·       A pressure of 1 atm or 100kilopascals (10²kPa)

·       A temperature of 298K (25^oC)

·       Reactants and products in physical states, normal for the above conditions.

·       A concentration of 1.0mol dm^-3 for solutions.

The enthalpy change of combustion will always have a negative value, of couse, because burning always releases heat. 

The standard enthalpy of combustion can be denoted as ΔH^o_c  where the ^o sign indicates standard conditions

When glucose burn in the oxygen it form carbon dioxide and water. So the proper equation of combustion of glucose is:

So the standard enthalpy of combustion will be:

∆H_combustion = (6 x ∆H_f CO₂ + 6 x ∆H_f H₂O) - ∆H_f C₆H₁₂O₆

                      = (6 x -400 + 6 x -300) – (-1300)

                         = -2900 kJ /mol

 The molar mass of glucose is 180 grams/mole. So

∆H_combustion = -2900/180 kJ/g

                   = -16.11 kJ/g

Thus, the standard enthalpy of combustion of per gram glucose at 25^oC is -16.11 kJ/g.  

Find the rate of reaction

Question: KI in acetone undergoes SN² reaction with each of P, Q, R and S. The rates of reaction vary as:

a)    P > Q > R > S

b)   S > P > R > Q

c)    P > R > Q > S

d)   R > P > S > Q

Solution: P,Q, R and S are alkyl halide. When any of them reacts with KI in the presence of acetone, alkyl iodide is formed. This is also known as Finkelstein reaction.

 In this reaction (I-) which is a nucleophile, replaces (Cl-), thus it is known as a nucleophilic substitution reaction. Since, in the reaction one bond is broken and one bond is formed in one step, it is called SN2 reaction (substitution nucleophilic bimolecular reaction).

The rate of SN2 reaction involves an interaction between nucleophile and the organic substrate. It can be described as:

rate = k [Nu][R-LG]. 

Where, Nu is nucleophile which is I- while [R-LG] is RX which are P, Q, R and S.

The relative reactivity for SN2 reaction in the given structure is:

According to this the correct option is (b). 

Find the radius of octahedral void

Question: The arrangement of X^- ions around A⁺ ion in solid AX is given in the figure (not drawn to scale). If the radius of X^- is 250pm, what is the radius of A⁺?

 

Solution:

According to given figure, A⁺ is present in the octahedral void of X-. An octahedral void is formed by six spheres which are arranged on the corner of two triangles. When these two triangles are placed with their apex faced opposite sides, octahedral void arises. 

The limiting radius in octahedral void is related to the radius of sphere as:

r_void = 0.414 r_sphere

                                        r _A+ = 0.414  r_X-

                                                   = 0.414 x 250 pm

                                                   = 103.5 pm

Thus, radius of A⁺ is 103.5 pm. 

Find the magnetic moment of the complex

Question: Consider the following complex ions, P, Q and R 

P = [FeF₆]^3-, Q = [V(H₂O)₆]²⁺, and R = [Fe(H₂O)₆]²⁺

What is the correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.)?

A) R<Q<P

B)  Q<R<P

C)  R<P<Q

D) Q<P<R

Solution:

The formula used to calculate the spin-only magnetic moment can be written as:

μ= √n(n+2) B.M.

Where n = no. of unpaired electron

 

So first let us find out number of unpaired electron in each complex.

 

1)   P = [FeF₆]^3-

Charge on Fe in the complex: [FeF₆]^3-    

                                                                                     X + (1x6) = -3

                                                        = +3

So number of unpaired electron in Fe⁺³ = 3s², 3p⁶, 3d⁵, 4s⁰            

                                                                      = 5

Magnetic moment = √5 (5+2)

                                         =√35 B.M.

 

2)   Q = [V(H₂O)₆]²⁺

Charge on Fe in the complex: X + (0x6) = +2

                                                 = +2

So number of unpaired electron in V⁺² = 3s², 3p⁶, 3d³,           

                                                                     = 3

Magnetic moment = √3 (3+2)

                                         =√15 B.M.

 

3)   R = [Fe(H₂O)₆]²⁺

Charge on Fe in the complex: X + (0x6) = +2

                                                    = +2

So number of unpaired electron in Fe⁺² = 3s², 3p⁶, 3d⁶         

                                                                     = 4

Magnetic moment = √4 (4+2)

                                         =√24 B.M.

So the correct order of the complex ions, according to their spin-only magnetic moment values is P > R > Q

Thus option B is correct. 

Racemises of 1- Chloro -1- phenylethane

Question:  A solution of (−)−1 − chloro − 1− phenylethane in toluene racemises slowly in the presence of a small amount of SbCl₅, due to the formation of :

(1) carbanion

(2) carbene

(3) carbocation

(4) free radical

Answer:  According to question, (−)−1 − chloro − 1− phenylethane racemises in the presence of SbCl_5.

 It  racemises slowly due to formation of intermediate compound.

 This intermediate compound is a carbocation. Thus, option c is correct. 

Colour of Nitric acid

Question: Why does concentrated nitric acid trun yellow brown upon long standing?

Answer: Nitric acid (HNO₃) is colourless in pure form. But it tends to acquire yellow brown colour as it become older. This is because it is a highly strong mineral acid. Due to this, it decomposes easily into oxides of nitrogen (NO₂), oxygen and water.

The produced nitrogen oxides dissolve partly or completely in the acid and give yellow or even red colour at higher temperatures. 

Find the total number of alkene

Question: How many numbers of alkenes possible by dehydrobromination of 3-bromo 3-cyclopentylhexane using alcoholic KOH?

Solution:

When any alkyl halide is treated with alcoholic KOH, it removes hydrogen and a halogen atom from it and forms alkene. This is basically a dehydrohalogenation reaction. (You can recall it’s basic from here). The halogen atom removes from the α-carbon and hydrogen atom from β-carbon of the alkyl halide.

 In 3-bromo 3-cyclopentylhexane, α-carbon atom is at position 3 (to which halogen atom attached) and three β-carbon atom which are at 2, 4 and 3 (C of cyclopentane). 

Because of three types of β-hydrogen, there are more than one alkenes are expected.

So let us find out.

11)   Hydrogen remove from cyclopentane i.e. position 3:

2

  2   2) Hydrogen remove from position 2:

Here two alkenes are obtained, one is cis from and other one is trans.

  3) Hydrogen remove from position 4:

Here also two alkenes are obtained, one is cis from and other one is trans.

So the total numbers of alkenes obtained from dehydrobromination of 3-bromo 3- cyclopentylhexane are 5. 

Find the number of electron

Question: What is the maximum number of electron that can have principal quantum number n = 3 and spin quantum number; m = -1/2?

Solution: The principal quantum number represents the main shells to which the electron belongs. 

We can calculate the number of electron in each shell by the formula: 2n²

Where n represent number of shell.

Here principal quantum number i.e. n is 3.

Thus, number of electron in the shell is 2(3)²

                                                        = 18

So the maximum number of electron that have principal quantum number 3 is 18.

Spin quantum number accounts for spin of electron.  An electron can spin clockwise or anticlockwise, in two opposite directions. For clockwise spin value is +1/2 and for anticlockwise spin value is -1/2.  

Generally in each orbital half electron spin clockwise direction and half in anti-clockwise direction.

So the maximum number of electron that show spin quantum number -1/2 will be 9 in principal quantum number 3. 

Find the optical isomerism of coordinate complex

Question: Which of the following complex species do not show optical isomerism?

a)     [Co(en)₃]³⁺

b)    [Co(en)₂Cl₂]⁺

c)     [Co(NH₃)₃Cl₃]

 

Solution: Lets us recall the basic of Optical Isomerism

a)      [Co(en)₃]³⁺

In this complex Co is central metal while en (ethylene diammine) is bidentate chelating ligand which is coordinate to the central metal atom Co. [Co(en)₃]³⁺ shows d²sp³ hybridisation.

Hybridisation of [Co(en)₃]³⁺:

 

This means that [Co(en)₃]³⁺  is an octahedral complex.

Such complex does not possesses any element of symmetry thus, it is optically active. It has been resolved into d- and l- forms.

 

b)    [Co(en)₂Cl₂]⁺

In this complex, two symmetrical bidentate chelating ligands ‘en’ (ethylene diammine) and two monodentate ligands ‘chlorine’ are coordinate to central metal atom Co.

[Co(en)₂Cl₂]⁺ shows d²sp³ hybridisation. This means that it is an octahedral complex.

Hybridisation of [Co(en)₂Cl₂]⁺ :

 

 [Co(en)₂Cl₂]⁺ exhibit the phenomenon of optical isomerism and can be resolved into their optical isomers. Its cis form is unsymmetrical, while the trans form is symmetrical because it contain plane of symmetry.

 Hence, optical isomerism is shown by cis form only. The cis form is resolved into d- and l- forms.

 

 c)     [Co(NH₃)₃Cl₃]

In this complex, Co is attached to three monodentate ligand ammonia and three (other) monodentate ligands chlorine.

 [Co(NH₃)₃Cl₃] shows d²sp³ hybridisation. This means that it is an octahedral complex.

Hybridisation of [Co(NH₃)₃Cl₃]:

In such octahedral complex, metal is placed at the centre and six ligands occupy their position at the vertices. Due to different spatial arrangement of the groups around the central metal, these complex show geometrical isomerism. This gives rise to two types of isomers called cis and trans isomers.

The cis and trans forms of [Co(NH₃)₃Cl₃] are given below.

 

 

 

In the cis-isomer, the three Cl- are one triangular face (position 1, 2, and 3) and the three NH₃ molecules are placed on the opposite triangular face (position 4, 5 and 6). Since three identical ligands occupy one face of an octahedron, the isomer is said to be facial or fac. 

In the trans isomer, Cl- are placed on the position 1, 2 and 3 while NH2 molecules are present on the positions 3, 4 and 5. It is also called meridional isomer (in which each set of three identical ligands occupies a plane passing through the metal atom).

[Co(NH₃)₃Cl₃] show facial as well as meridional isomerism. But both contain plane of symmetry, hence it is optically inactive.

 

 

 

Find the magnetic property of complex

Question: Which of the following complexes are diamagnetic in nature?

a)     K₃[Fe(CN)₆]

b)    [Co(NH₃)₆]Cl₃

c)     Na₃[Co(Oxalate)₃]

d)    [NiCl₄]^-2

Solution:

Magnetic property of a complex depends upon the number of unpaired electrons present in the orbital of its metal.

·        If unpaired electron present in the complex then it is paramagnetic in nature.

·        If there is no unpaired electron then the complex is diamagnetic in nature.

Now let us start with each option and find out its magnetic property.

a)     K₃[Fe(CN)₆]

It is a anionic complex because it carries negative charge on complex ion.

                                                                      (Complex Ion)

In this complex ion Fe is central metal and CN is ligand. To calculate number of electron in Fe we have to know its oxidation state. Since the whole complex ion show (-3) charge so the oxidation number of Fe will be:

Charge number of [Fe(CN)₆]^3-  = charge on Fe + charge on CN

                                                 -3 = x + 6 (-1)     (CN has -1 charge)

                                                  X = +3

Now let us find out number of electrons in the central atom Fe³⁺

Fe (26) = 2, 8, 14, 2

           = 3s², 3p⁶, 3d⁶, 4s²

 

 Fe³⁺ = 3s², 3p⁶, 3d⁵, 4s⁰            

 

In the presence of CN as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes one electron unpaired in 3d.

Due to presence of this unpaired electron, K₃[Fe(CN)₆] shows paramagnetic.

 

b)    [Co(NH₃)₆]Cl₃

It is a cationic complex because it carries positive charge on complex ion.

In this complex ion Co is central metal and NH₃ is ligand. Now let us know the oxidation number of Co:

Charge number of [Co(NH₃)₆]³⁺  = charge on Co + charge on NH₃

                                                  +3 = x + 6 (0)     (NH₃ is neutral)

                                                  X = +3

Now let us find out number of electrons in the central atom Co³⁺

Co (27) = 3s², 3p⁶, 3d⁷, 4s²

 

  Co³⁺ = 3s², 3p⁶, 3d⁶, 4s⁰            

 

In the presence of NH₃ as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes all electron paired in 3d.

Hence, [Co(NH₃)₆]Cl₃ shows diamagnetic.

 

c)     Na₃[Co(Oxalate)₃]

It is an anionic complex because it carries negative charge on complex ion.

In this complex ion Co is central metal and oxalate is ligand. Now let us know the oxidation number of Co:

Charge number of [Co(oxalate)₃]^3-  = charge on Co + charge on NH₃

                                                           -3 = x + 6 (-1)     (oxalate, C₂O₄, has -1 charge)

                                                             X = +3

Now let us find out number of electrons in the central atom Co³⁺

Co (27) = 3s², 3p⁶, 3d⁷, 4s²

  Co³⁺ = 3s², 3p⁶, 3d⁶, 4s⁰            

 

In the presence of oxalate (C₂O₄) as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes all electron paired in 3d.

Hence, Na₃[Co(oxalate)₃] shows diamagnetic.

 d)     [NiCl₄]^-2

In this complex ion Ni is central metal and Cl is ligand. Now let us know the oxidation number of Ni:

Charge number of [NiCl₄]^2-  = charge on Ni + charge on H₂O

                                                  -2 = x + 4 (-1)     (Cl has -1 oxidation number)

                                                  X = +2

Now let us find out number of electrons in the central atom Ni²⁺

Ni (28) = 3s², 3p⁶, 3d⁸, 4s²

 

  Ni²⁺ = 3s², 3p⁶, 3d⁸, 4s⁰            

In the presence of Cl as ligand, one 4s and three 4p orbitals undergo sp3 hybridisation.  Each Cl- ion donates a pair of electrons. But this makes two electrons unpaired in 3d.

Hence, [NiCl₄] ^2- shows paramagnetic nature.

Thus, options b and c are diamagnetic in nature.