Chemistry Test - 17

In the solid state, hydrogen fluoride exists as an infinite zig-zag chain of molecules.

When a metal reacts with oxygen metallic oxide is formed.

Many metals react with oxygen to form metal oxides. Potassium and sodium are soft metals that are easily cut exposing a shiny surface that changes to dull rapidly.The change from shiny to dull is called tarnishing.

For example:

• Potassium burns with a lilac flame when heated in the air.
• Potassium \(+\) Oxygen \(\rightarrow\) Potassium oxide
• \(4 {~K}({~s})+{O}_{2}({~g}) \rightarrow 2 {~K}_{2} {O}({s})\)
• Magnesium reacts readily in the air burning with white light.
• Magnesium \(+\) Oxygen \(\rightarrow\) Magnesium oxide
• \(2 {Mg}({s})+{O}_{2}({~g}) \rightarrow 2 {Mg} {O}({s})\)

The common name for 4–hydroxy toluene is p–cresol.

AgBr has highest solubility in 10-3 M NH4OH. AgBr forms a soluble complex in with NH4OH, i.e. [Ag(NH3)2]Br.

Magnetic moment depends upon the number of unpaired electrons :

(a)$V^{4+}\left(Z=23\right):\left(Ar\right)3d^1:1$unpaired electron

(b)$M{n}^{4+}\left(Z=25\right):\left(Ar\right)3d^3:3$unpaired electrons

(c)$F{e}^{3+}\left(Z=26\right):\left(Ar\right)3d^5:5$unpaired electrons

(d)$N{i}^{2+}\left(Z=28\right):\left(Ar\right)3d^8:2$unpaired electrons

Numbers of unpaired electrons are maximum in$F{e}^{3+}.$Hence, it shows maximum, magnetic moment and$V^{4+}$shows least magnetic moment because it has minimum number of unpaired electrons.

The order of magnetic moment is as :

$F{e}^{3+}>M{n}^{4+}>N{i}^{2+}>V^{4+}$
$\downarrow \downarrow \downarrow \downarrow$
(c) (d) (d) (a)

\({C H}_{2}={C H}-{C H}_{2}-{C l}\) is known as allyl chloride.

Allyl chloride has a parent chain with three carbons and a double bond. Therefore, the name of the parent chain will be propene. Numbering will start from the left side, double will get number one and substituent chlorine will get number three.

So, the IUPAC name is 3-chloro-1-propene.

K2Cr2O7 + 1H2SO4 + 3SO2K2SO4 + Cr2(SO4)3 + 1H2O

The correct order of the basicity is

$N{H_2}^{-}>R{O}^{-}>O{H}^{-}>RC{O}^{-}$

$N{H_2}^{-}$is most basic and$RCO{O}^{-}$is the lost