Chemistry Test 135

Boyle Temperature T_b= a/R_b

Critical Temperature T_c= 8a/27R_b

Dividing both

T_b/T_c =27/8

A reducing sugar is one which is capable of acting as reducing agent.
All monosaccharaides are reducing sugars because all monosaccharaides have an aldehyde group (if they are aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses) and can themselves get oxidized hence can act as reducing agent.

Galactose
All of them are monosaccharides.

Cis –isomer
Square planar complexes having symmetric bidentate ligands carrying one or more substituents can form geometrical isomers. Because, such ligand show kind of asymmetry. Here pn (propylenediamine) shows this kind of nature hence leading to geometrical isomerism.
All the rest options are of symmetrical in nature around metal ion so no geometrical isomerism is possible in these.

Volume of unit cell = (288 pm)³ = (288 × 10^–12 m)³
= (288 × 10^–10 cm)³
=2.39 × 10^–23 cm³
For bcc, z_eff = 2

= 12.08 × 10²³ unit cells
Since each bcc cubic unit contains 2 atoms, therefore the total number of atoms in 208 g
= 2 (atoms/unit cell) × 12.08 × 10²³ unit cells
= 24.16 × 10²³ atoms

The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H₅IO₆ has an oxidation state of +7 and hence is the strongest acid.

Before we commence, we note that the spontaneous cell reaction is

a) H₂O →OH- + H⁺
NH₃→NH₂^- +H⁺
Water being more polar than ammonia dissociates more and gives plenty amount of H⁺ and hence H₂ gas is released as result. In ammonia sufficient amount of H⁺ is not obtained due to less polar nature (so less dissociation) hence H₂ gas is not evolved and then ammoniation of electron happens

b) As we dilute further dissociation of ammonia increases (because degree of dissociation is inversely proportional to concentration) and we get more H⁺ and NH₂^- ion. H⁺ combines with ammoniated electron and H₂ gas is evolved and paramagnetism is gone.

c) NH₃→NH₂^- +H⁺
M →M⁺ +e^-
M⁺ +xNH₂^-→[M (NH₂)x] type of complex
On adding d block metal ion we see that metal ion combines with NH₂^- and forms metal complex. Complex formation leads to ammonia dissociation even more and as a result of ammonia dissociation more H⁺ is released and which combines with ammoniated electron and H₂ gas is evolved. Now colour and magnetic nature may or may not change. But whatever it will be it will be due to this complex.

d) Metallic bonding increases at very high concentration M₂ is formed.

Condition for ‘ligand to metal charge transfer’ are

i) Metal should be in high oxidation state so that it has high ionization energy. And it should also be of smaller size with vacant orbitals having low energies.

ii) Ligand should have lone pair of electrons having high energy
In KMnO_4, Mn is in +7 oxidation state and have all the 3d orbitals vacant. Mn⁺⁷ ion is surrounded by four oxide ions. All oxide ions have filled 2p orbitals. There is transfer of an electron from filled 2p orbital of oxide ion to vacant d orbitals of Mn⁺⁷ ion. Due to this transfer of electron from oxides of KMnO₄ to metal happens and as a result colouris intensely purple.

Trick: Almost all the metal ion of d block having d0 configuration tend to show colour due to ligand to metal charge transfer. Another example of this is K₂Cr₂O₇

In water the product is almost all benzyl naphthol. However, in DMSO (dimethyl sulfoxide) the major product is the ether. In water the oxyanion is heavily solvated through hydrogen bonds to water molecules and the electrophile cannot push them aside to get close to O– (this is an entropy effect). DMSO cannot form hydrogen bonds as it has no OH bonds and does not solvate the oxyanion, which is free to attack the electrophile.

DMSO Solvent:

Correct order is
benzene > thiophene > pyrrole > furan.
Take a look at the structures

Electronegativity of atoms in the ring
O>N>S
In furan oxygen is in ring. It will be less reluctant to share electrons of lone pair for resonance and hence aromatic character will be least in it.
In pyrrole nitrogen is in ring. It will be more willing than furan to share electrons for resonance hence more aromatic than furan.
Thiophene has sulphur which is less electronegative than nitrogen hence more aromatic character than pyrrole.
Benzen has carbon and has 6π electrons already so most aromatic.