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use the table of electronegativities, above, to determine which type of…

Question

use the table of electronegativities, above, to determine which type of bond should be formed between the two atoms.

  1. phosphorus silicon
  1. cobalt bromine
  1. germanium selenium
  1. silicon fluorine
  1. potassium nitrogen
  1. cesium oxide
  1. hydrogen phosphorus
  1. arsenic chloride
  1. lithium chloride
  1. iron carbon

Explanation:

Step1: Find electronegativity values

From the table:

  • Phosphorus ($P$) has electronegativity $2.1$, Silicon ($Si$) has $1.8$.
  • Cobalt ($Co$) has $1.9$, Bromine ($Br$) has $3.0$.
  • Germanium ($Ge$) has $1.8$, Selenium ($Se$) has $2.5$.
  • Silicon ($Si$) has $1.8$, Fluorine ($F$) has $4.0$.
  • Potassium ($K$) has $0.8$, Nitrogen ($N$) has $3.0$.
  • Cesium ($Cs$) has $0.7$, Oxygen ($O$) has $3.5$.
  • Hydrogen ($H$) has $2.1$, Phosphorus ($P$) has $2.1$.
  • Arsenic ($As$) has $2.1$, Chlorine ($Cl$) has $3.0$.
  • Lithium ($Li$) has $1.0$, Chlorine ($Cl$) has $3.0$.
  • Iron ($Fe$) has $1.8$, Carbon ($C$) has $2.5$.

Step2: Calculate electronegativity differences

  • For $P$ and $Si$: $|2.1 - 1.8|=0.3$.
  • For $Co$ and $Br$: $|3.0 - 1.9| = 1.1$.
  • For $Ge$ and $Se$: $|2.5 - 1.8|=0.7$.
  • For $Si$ and $F$: $|4.0 - 1.8|=2.2$.
  • For $K$ and $N$: $|3.0 - 0.8|=2.2$.
  • For $Cs$ and $O$: $|3.5 - 0.7|=2.8$.
  • For $H$ and $P$: $|2.1 - 2.1|=0$.
  • For $As$ and $Cl$: $|3.0 - 2.1|=0.9$.
  • For $Li$ and $Cl$: $|3.0 - 1.0|=2.0$.
  • For $Fe$ and $C$: $|2.5 - 1.8|=0.7$.

Step3: Determine bond type

  • If $\Delta EN=0$: non - polar covalent.
  • If $0\lt\Delta EN\lt1.7$: polar covalent.
  • If $\Delta EN\geq1.7$: ionic.
  • $P$ and $Si$: $\Delta EN = 0.3$, polar.
  • $Co$ and $Br$: $\Delta EN=1.1$, polar.
  • $Ge$ and $Se$: $\Delta EN = 0.7$, polar.
  • $Si$ and $F$: $\Delta EN=2.2$, ionic.
  • $K$ and $N$: $\Delta EN=2.2$, ionic.
  • $Cs$ and $O$: $\Delta EN=2.8$, ionic.
  • $H$ and $P$: $\Delta EN = 0$, non - polar covalent.
  • $As$ and $Cl$: $\Delta EN=0.9$, polar.
  • $Li$ and $Cl$: $\Delta EN=2.0$, ionic.
  • $Fe$ and $C$: $\Delta EN = 0.7$, polar.

Answer:

Atom 1Atom 2Electronegativity Difference ValueBond Type
CobaltBromine$1.1$Polar
GermaniumSelenium$0.7$Polar
SiliconFluorine$2.2$Ionic
PotassiumNitrogen$2.2$Ionic
CesiumOxide$2.8$Ionic
HydrogenPhosphorus$0$Non - polar Covalent
ArsenicChloride$0.9$Polar
LithiumChloride$2.0$Ionic
IronCarbon$0.7$Polar