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blsea [12.9K]
3 years ago
14

What is the electron configuration of a chlorine ion in a compound of BeCl2?

Chemistry
2 answers:
mestny [16]3 years ago
8 0
The  electronic  configuration  of  a   chlorine ion  in BeCl2  compound is

[2.8.8]^-  (answer  B)

chlorine  atom  gain  on electron form Be to  form  chloride ions
chlorine  atom  has  a  electronic   configuration of  2.8.7   and  it  gains one  electron  to  form  chloride ion  with  2.8.8  electronic  configuration
Brrunno [24]3 years ago
5 0
Chlorine has 7 valence electrons in the outermost shell. when it forms an ionic bond with Be, it takes in 1 electron and achieves the noble gas configuration
since it has one more electron it gains a negative charge of -1.
anion is then - Cl⁻
electronic configuration of Cl atom is - [2,8,7]
after taking in the electron Cl⁻ - [2,8.8]⁻
answer is 
<span>B. [2.8.8]</span>⁻
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The densities of gases a, b, and c at stp are 1.25 g/l, 2.86 g/l and 0.714 g/l, respectively. calculate the molar mass of each s
kicyunya [14]
The  molar  mass  of    a, b and  c at  STP is calculated  as  below

At  STP  T  is always=   273 Kelvin and ,P= 1.0 atm 

by  use of  ideal  gas  equation  that  is  PV =nRT
n(number   of moles) = mass/molar mass  therefore  replace   n  in  the  ideal   gas  equation

that  is Pv = (mass/molar mass)RT
multiply  both side  by molar  mass  and  then  divide  by  Pv  to  make  molar mass the  subject of the  formula

that is  molar  mass =  (mass x RT)/ PV

 density is always = mass/volume

therefore  by  replacing  mass/volume  in   the equation  by  density the equation
molar  mass=( density  xRT)/P  where R  =  0.082 L.atm/mol.K

the  molar mass  for  a
= (1.25 g/l  x0.082 L.atm/mol.k  x273k)/1.0atm = 28g/mol

the molar  mass of b
=(2.86g/l  x0.082L.atm/mol.k   x273  k) /1.0  atm  = 64  g/mol

the molar  mass of c

=0.714g/l  x0.082  L.atm/mol.K  x273 K) 1.0atm= 16 g/mol

therefore  the 
   gas  a  is  nitrogen N2   since 14 x2= 28 g/mol
   gas b =SO2  since  32 +(16x2)= 64g/mol
  gas c =   methaneCH4  since  12+(1x4) = 16 g/mol


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3 years ago
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3 years ago
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zhuklara [117]

Answer:

  • <u>Cadmium has larger atomic radius than sulfur.</u>

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

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  • Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

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So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

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Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

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