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garri49 [273]
3 years ago
14

Choose the selection which correctly characterizes all three of the following substances in terms of whether they are polar or n

onpolar: SiH4 and BBr3 and SiF4 a) SiH4 is nonpolar and BBr3 is polar and SiF4 is nonpolar. b) SiH4 is nonpolar and BBr3 is polar and SiF4 is polar. c) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is polar. d) SiH4 is polar and BBr3 is nonpolar and SiF4 is polar. e) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.
Chemistry
1 answer:
ohaa [14]3 years ago
4 0

Answer:

SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.

Explanation:

SiH4 is a non-polar compound. Though the Si–H bonds are polar, as a result of different electronegativities of Si and H. However, as there are 4 electron repulsions around the central Si atom, the polar bonds are arranged symmetrically around the central atom having a tetrahedral shape hence they cancel out making the compound nonpolar.

SiF4 is a nonpolar molecule because the fluorine atoms are arranged symetrically around the central silicon atom in a tetrahedral molecule with all of the regions of negative charge cancelling each other out just like in SiH4.

The 3 bromine atoms all lie in the same plane thus the geometry of the compound will be trigonal planar. The BBr3 will be non polar because the three B-Br bonds will cancel out each others' dipole moment given that they are in the same plane.

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Answer:

+2

Explanation:

Alkaline earth metals are present in group 2 of periodic table. There are six elements in second group. Beryllium, magnesium, calcium, strontium, barium and radium.

All have two valance electrons.

Electronic configuration of Beryllium:

Be = [He] 2s²

Electronic configuration of magnesium.

Mg = [Ne] 3s²

Electronic configuration of calcium.

Ca = [Ar] 4s²

Electronic configuration of strontium.

Sr = [Kr] 5s²

Electronic configuration of barium.

Ba = [Xe] 6s²

Electronic configuration of radium.

Ra = [ Rn] 7s²

They are present in group two and have same number of valance electrons (two valance electrons) and show oxidation state +2 by loosing two valance electrons. They also show similar reactivity.

They react with oxygen and form oxide.

2Ba   +   O₂   →    2BaO

2Mg  +   O₂   →    2MgO

2Ca +   O₂   →    2CaO

this oxide form hydroxide when react with water,

BaO  + H₂O   →  Ba(OH)₂

MgO  + H₂O   →  Mg(OH)₂

CaO  + H₂O   →  Ca(OH)₂

With sulfur,

Mg + S   →  MgS

Ca + S   →  CaS

Ba + S   →  BaS

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So the dissolution equation of NaI is:
NaI(s) → Na^+(s)  +   I^-(Aqu) 
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9. A student observed the following reaction:
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Answer:

The substance that remained on the filter paper is Al(OH).

Explanation:

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