They form blue coloured solution. The solution is paramagnetic in nature.

711\(\mu\)

The atomic mass of each He is 4.003, to be precise. That means the group of each atom is 4.003 amu. Then the 52 u of He has 52/4.003, which is 12.99 means 13. So, 52 u of He has 13 atoms.

Wurtz synthesis: Higher alkanes are prepared by heating an alkyl halide (RX) with sodium metal in dry ether solution.
R - X + 2Na + XR \(\overset { ether }{ \longrightarrow } \) R - R + 2NaX
CH₃Br + 2Na + BrCH_{3\(\longrightarrow \)} CH₃-CH₃ + 2NaBr
Limitations: Use of two different alkyl halides in Wurtz reaction always leads to a mixture of alkanes. The separation of these alkanes is difficult because there is only a little difference in their boiling points. Thus only symmetrical alkanes can be prepared by this method.

Diamond	Graphite
Diamond is the hardest substance on earth.	Graphite is soft and slippery
In diamond carbon is Sp3- hybridized	In Graphite carbon is Sp2- hybridized
Since all the electrons in diamond are firmly held in C-C,6 bonds there are no free electrons in diamond crystal Therefore diamond is bad conductor of electricity	Since only three electrons of each carbon are used in making hexagonal rings of graphite, fourth valence electron is free to move thus graphite is a good conductor of electricity
Because of high refractive index diamond can reflect and refract the light.	Graphite is a black substance and possess a metallic lustre

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(a)	Lithium	Group 1	1	1	\({ Li }_{ 2 }O\)
	Oxygen	Group 16	6	8 - 6 = 2

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(b)	Group 13	Group 2	2	2	\({ Mg }_{ 3 }{ N }_{ 2 }\)
	Group 17	Group 15	5	8 - 5 = 3

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(c)	Group 13		3	3	\({ AI I}_{ 3 }\)
	Group 17		7	8 - 7 = 1

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(d)	Group 14		4	4	\({ SiO }_{ 2 }\)
			6	8 - 6 = 2

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(e)	Group 15		5	3 or 5	PF3 or PF5.
	Group 17		7	8 - 7 = 1

S.No	Element	Group number	Electrons in valence shell	Valency	Formulae of binary compound
(f)	Group 3	Group 3	3	3	\({ LuF }_{ 3 }\)
	Group 17	Group 17	7	8 - 7 = 1

a. Lithium is an alkali metal (Group1). It has only one electron in the valence shell, therefore, its valence is 1.Oxygen is a group 16 element with a valence of 2. Therefore, formula of the compound formed would be Li₂O (lithium oxide).
b. Magnesium is an alkali earth metal (Group2). hence has a valence is 2.Nitrogen is a group 15 element with a valence of 8−5=3. Thus, the formula of the compound formed would be Mg₃N₂ (magnesium nitride).
c. Aluminium is group 13 elements with a valence of 3 while iodine is a halogen (group17) with a valence of 1. Therefore, the formula of the compound formed be AII₃(Aluminium iodide.).
d. Silicon is group 14 elements with a valence of 4 while oxygen is a (group16) with a valence of (8−6=2). Hence, the formula of the compound formed be SiO₂(silicon dioxide).
e. Phosphorus is a group 15 element with a valence of 3 or 5 while flurine is group 17 element with a valence of 1. Hence, the formula of the compound formed would be PF₃ or PF₅.
f. Element with atomic number 71 is a lanthanoid called lutetium (Lu). Its common valence is 3. Fluorine is a group 17 (halogen) element with a valence of 1. Therefore, the formula of the compound formed would be LuF₃(lutetium fluoride).

1 mole of NH₃ =10 moles of protons
= \(6.022\times 10^{ 23 }\times 10 \ protons\)
1 mole of NH₃ (or 17g) contains 6.022 x 10²⁴ protons
34 mg or 34 x 10^-3 g NH₃ will contain
\(=\frac { 34\times { 10 }^{ -3 }\times 6.022\times { 10 }^{ 24 } }{ 17 }\)
\(=12.044\times 10^{ 21 } \ protons\)
\(=1.2044\times 10^{ 22 } \ protons\)
Mass of 1 proton = 1.6726x10^-27kg
Mass of 1.2044 x 10²² x 1.6726 x 10^-27kg
= 2.01447 x 10^-5kg

Number of moles of HCI
= 250 mL\(\times \frac { 0.76M }{ 1000 } =0.19 \ mol;\)
Number of moles of CaCo₃ = \(\frac { 1000g }{ 100 \ g \ { mol }^{ -1 } } \)= 10 mol
For the 10 moles of CaCO₃(s) number of moles of HCI
required would be \(10\times \frac { 2 }{ 1 } \) = 20 mol HCI (aq)
But we have only 0.19 mole HCI (aq), hence, HCI(aq) is the limiting reagent. Since, 2 moles HCI (aq) forms 1 mole
of CaCl₂, therefore, 0.19 moles of HCI (aq) would give
\(0.19\times \frac { 1 }{ 2 } =0.095 \ mol\)
Mass of CaCI₂ = 0.095 \(\times\) 111 = 10.54 g