Clarify scope of breaking change (#22012)

Co-authored-by: Genevieve Warren <[email protected]>

Author: tdykstra

docs/core/compatibility/serialization/5.0/deserializing-json-into-char-requires-single-character.md

@@ -1,25 +1,27 @@
 ---
-title: "Breaking change: Deserialize requires single-character string"
-description: Learn about the breaking change in .NET 5.0 where JsonSerializer.Deserialize requires a single-character string.
-ms.date: 10/18/2020
+title: "Breaking change: Deserialize char requires single-character string"
+description: Learn about the breaking change in .NET 5.0 where System.Text.Json requires a single-char string in the JSON when deserializing to a char target.
+ms.date: 12/15/2020
 ---
-# JsonSerializer.Deserialize requires single-character string
+# System.Text.Json requires single-char string to deserialize a char
 
-When the type parameter is <xref:System.Char>, the string argument to <xref:System.Text.Json.JsonSerializer.Deserialize%60%601(System.String,System.Text.Json.JsonSerializerOptions)?displayProperty=nameWithType> must contain a single character for deserialization to succeed.
+To successfully deserialize a <xref:System.Char> using <xref:System.Text.Json>, the JSON string must contain a single character.
 
 ## Change description
 
-In previous .NET versions, if you pass a multi-character string to <xref:System.Text.Json.JsonSerializer.Deserialize%60%601(System.String,System.Text.Json.JsonSerializerOptions)?displayProperty=nameWithType> and the type parameter is <xref:System.Char>, the deserialization succeeds and only the first character is deserialized.
-
-In .NET 5.0 and later, when the type parameter is <xref:System.Char>, passing anything other than a single-character string causes a <xref:System.Text.Json.JsonException> to be thrown.
+In previous .NET versions, a multi-`char` string in the JSON is successfully deserialized to a `char` property or field. Only the first `char` of the string is used, as in the following example:
 
 ```csharp
-// .NET Core 3.0 and 3.1: Returns the first character 'a'.
-// .NET 5.0 and later: Throws JsonException because payload has more than one character.
-JsonSerializer.Deserialize<char>("\"abc\"");
+// .NET Core 3.0 and 3.1: Returns the first char 'a'.
+// .NET 5.0 and later: Throws JsonException because payload has more than one char.
+char deserializedChar = JsonSerializer.Deserialize<char>("\"abc\"");
+```
 
+In .NET 5.0 and later, anything other than a single-`char` string causes a <xref:System.Text.Json.JsonException> to be thrown when the deserialization target is a `char`. The following example string is successfully deserialized in all .NET versions:
+
+```csharp
 // Correct usage.
-JsonSerializer.Deserialize<char>("\"a\"");
+char deserializedChar = JsonSerializer.Deserialize<char>("\"a\"");
 ```
 
 ## Version introduced
@@ -28,21 +30,21 @@ JsonSerializer.Deserialize<char>("\"a\"");
 
 ## Reason for change
 
-<xref:System.Text.Json.JsonSerializer.Deserialize%60%601(System.String,System.Text.Json.JsonSerializerOptions)?displayProperty=nameWithType> parses text that represents a single JSON value into an instance of the type specified by the generic type parameter. Parsing should only succeed when the provided payload is valid for the specified generic type parameter. For a <xref:System.Char> value type, a valid payload is a single character string.
+Parsing for deserialization should only succeed when the provided payload is valid for the target type. For a `char` type, the only valid payload is a single-`char` string.
 
 ## Recommended action
 
-When parsing a string into a <xref:System.Char> type using <xref:System.Text.Json.JsonSerializer.Deserialize%60%601(System.String,System.Text.Json.JsonSerializerOptions)?displayProperty=nameWithType>, make sure the string consists of a single character.
+When you deserialize JSON into a `char` target, make sure the string consists of a single `char`.
 
 ## Affected APIs
 
-- <xref:System.Text.Json.JsonSerializer.Deserialize%60%601(System.String,System.Text.Json.JsonSerializerOptions)?displayProperty=fullName>
+- <xref:System.Text.Json.JsonSerializer.Deserialize%2A?displayProperty=fullName>
 
 <!--
 
 ### Affected APIs
 
-- `M:System.Text.Json.JsonSerializer.Deserialize``1(System.String,System.Text.Json.JsonSerializerOptions)`
+- `Overload:System.Text.Json.JsonSerializer.Deserialize`
 
 ### Category