Answer: Reducing agent in the given reaction is
.
Explanation:
A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.
In the given reaction, oxidation state of sulfur in
is +2 and
has 0 oxidation state.
In
oxidation state of S is 2.5 and in
oxidation state of I is -1.
Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.
Thus, we can conclude that reducing agent in the given reaction is
.
Answer:
Lewis structure in attachment.
Explanation:
Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.
The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.
Answer:
see explanation
Explanation:
Write the balanced COMPLETE ionic equation for the reaction when Na₂CO₃ and AgNO₃ are mixed in aqueous solution. If no reaction occurs, simply write only NR.
Ag (+1) + NO3(-1) + 2 Na(+1) + Co3 (-2)--> Ag2CO3 (s) + 2 Na (+1) + 2NO3(-1)
we are given the the two reactants: AgNO3 and Na2CO3 and is asked to write a balanced equation and a net ionic equation for the reaction of the two. This is a double-replacement reaction:
2AgNO3 (aq)+ Na2CO3 (aq)= Ag2CO3 + 2NaNo3 (aq)
2 Ag + + 2 N03- + 2Na+ + CO32- = Ag2CO3 + 2 Na+ 2NO3-
cancelling the spectator ions, 2Ag + + CO32- = Ag2CO3